Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 12 Distance And Section Formulae

Exercise 12.1

Answer 2. Find the distance between the following pairs of points:
Answer: (a) P(0,0), Q(5,12)
Coordinates of origin are P(0, 0).
\( PQ = \sqrt{(12 - 0)^2 + (5 - 0)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ units} \) (b) P(0,0), Q(6,8)
\( PQ = \sqrt{(6 - 0)^2 + (8 - 0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ units} \) (c) P(0,0), Q(8,15)
\( PQ = \sqrt{(8 - 0)^2 + (15 - 0)^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \text{ units} \) (d) P(0,0), Q(0,11)
\( PQ = \sqrt{(0 - 0)^2 + (11 - 0)^2} = \sqrt{121} = 11 \text{ units} \) (e) P(0,0), Q(13,0)
\( PQ = \sqrt{(13 - 0)^2 + (0 - 0)^2} = \sqrt{169} = 13 \text{ units} \)
In simple words: We use the distance formula to find how far apart two points are. It is like finding the straight line distance between two places.

📝 Teacher's Note: Draw points on graph paper to show students the distance formula is like finding the hypotenuse of a right triangle. Use the Pythagorean theorem which students already know.

🎯 Exam Tip: Always write the distance formula first. Then substitute the coordinates carefully. Check your arithmetic twice when finding square roots.

Answer 3. Find the distance between the following pairs of points:
Answer: (a) A(p+q, p-q), B(p-q, p-q)
\( AB = \sqrt{(p - q - p - q)^2 + (p - q - p + q)^2} \)
\( = \sqrt{4q^2 + 0} = 2q \text{ units} \) (b) A(sinθ, cosθ), B(cosθ, -sinθ)
\( AB = \sqrt{(cosθ - sinθ)^2 + (-sinθ - cosθ)^2} \)
\( = \sqrt{cos^2θ + sin^2θ - 2cosθsinθ + sin^2θ + cos^2θ + 2cosθsinθ} \)
\( = \sqrt{2} \text{ units} \) (c) A(secθ, tanθ), B(-tanθ, secθ)
\( AB = \sqrt{(-tanθ - secθ)^2 + (secθ - tanθ)^2} \)
\( = \sqrt{tan^2θ + sec^2θ + 2tanθsecθ + sec^2θ + tan^2θ - 2tanθsecθ} \)
\( = \sqrt{2sec^2θ + 2tan^2θ} \text{ units} \) (d) A(sinθ - cosecθ, cosθ - cotθ), B(cosθ - cosecθ, -sinθ - cotθ)
\( AB = \sqrt{(cosθ - cosecθ - sinθ + cosecθ)^2 + (-sinθ - cotθ - cosθ + cotθ)^2} \)
\( = \sqrt{(cosθ - sinθ)^2 + (-sinθ - cosθ)^2} \)
\( = \sqrt{cos^2θ + sin^2θ - 2cosθsinθ + sin^2θ + cos^2θ + 2sinθcosθ} \)
\( = \sqrt{2} \text{ units} \)
In simple words: These problems use trigonometry (sine, cosine, etc.) but the distance formula works the same way. We just need to be careful with the algebra.

📝 Teacher's Note: Remind students that sin²θ + cos²θ = 1 always. This identity helps simplify many trigonometric distance problems quickly.

🎯 Exam Tip: Write each step clearly when expanding brackets. Use trigonometric identities to simplify. Always check if your final answer makes sense.

Answer 4. The abscissa of a point is -5 and the distance from the point (7,5) is 13. Find the ordinates of the point.
Answer: Let the point on x-axis be (x, 0) given abscissa is -5.
∴ point is P(-5, 0)
Let (7, 5) be point A
\( AP = \sqrt{(7 + 5)^2 + (5 - 0)^2} \)
\( = \sqrt{144 + 25} \)
\( = \sqrt{169} \)
\( = 13 \text{ units} \)
In simple words: The abscissa is the x-coordinate. We know x = -5 and the distance to another point is 13. We use the distance formula to find the y-coordinate.

📝 Teacher's Note: Explain that abscissa means x-coordinate and ordinate means y-coordinate. Use a graph to show these terms clearly.

🎯 Exam Tip: Always write "Let the point be (x,y)" first. Substitute known values into the distance formula. Show all algebraic steps to find the unknown coordinate.

Answer 5. A point lies on the line y = 0. Its abscissa is 1 and its distance from the point (13, -9) is 15. Find the point.
Answer: Point on the line y = 0 lies on x-axis given abscissa is 1.
∴ point is P(1, 0)
Let (13, -9) be point A
\( AP = \sqrt{(13 - 1)^2 + (-9 - 0)^2} \)
\( = \sqrt{144 + 81} \)
\( = \sqrt{225} \)
\( = 15 \text{ units} \)
In simple words: The line y = 0 is the x-axis. So our point has y-coordinate = 0. We know x = 1, so the point is (1, 0).

📝 Teacher's Note: Show students that y = 0 means the x-axis. Any point on the x-axis has its y-coordinate as zero.

🎯 Exam Tip: When a point lies on y = 0, it means y-coordinate is 0. When it lies on x = 0, it means x-coordinate is 0. Remember this rule.

Answer 6. A point lies on the line x = 0. Its ordinate is 9 and its distance from the point (12, 5) is 4√10. Find the point.
Answer: Point on the line x = 0 lies on given its ordinate is 9.
∴ point is P(0, 9)
Let the point (12, 5) be A.
\( AP = \sqrt{(12 - 0)^2 + (5 - 9)^2} \)
\( = \sqrt{144 + 16} \)
\( = \sqrt{160} \)
\( = 4\sqrt{10} \text{ units} \)
In simple words: The line x = 0 is the y-axis. So our point has x-coordinate = 0. We know y = 9, so the point is (0, 9).

📝 Teacher's Note: Explain that x = 0 means the y-axis. Any point on the y-axis has its x-coordinate as zero. Use a graph to show this clearly.

🎯 Exam Tip: Always identify which axis the point lies on first. Then substitute the known coordinates into the distance formula and solve for unknowns.

Answer 7. If the distance between the points (5, a) and (1, -5) is 5, find the value of a.
Answer: Let the points (5, a) and (1, -5) be P and Q respectively.
Given, PQ = 5 units
\( \sqrt{(5 - 1)^2 + (a + 5)^2} = 5 \) Squaring both sides, we get,
\( 16 + a^2 + 25 + 10a = 25 \)
\( \Rightarrow a^2 + 10a + 16 = 0 \)
\( \Rightarrow a^2 + 8a + 2a + 16 = 0 \)
\( \Rightarrow (a + 8)(a + 2) = 0 \)
∴ a = -8, -2
In simple words: We put the given distance value into the distance formula. Then we solve the equation to find the unknown coordinate a.

📝 Teacher's Note: Teach students to square both sides when dealing with square roots in equations. Show how to factorize the quadratic equation step by step.

🎯 Exam Tip: When distance is given, substitute into distance formula and square both sides. Solve the quadratic equation carefully. There may be two possible answers.

Answer 8. If the distance between the points (m, -4) and (3, 2) is 3√5, find the value of m.
Answer: Let the points (m, -4) and (3, 2) be A and B respectively.
Given AB = 3√5 units
\( \sqrt{(m - 3)^2 + (-4 - 2)^2} = 3\sqrt{5} \) Squaring both sides
\( m^2 - 6m + 9 + 36 = 45 \)
\( \Rightarrow m^2 - 6m = 0 \)
\( \Rightarrow m(m - 6) = 0 \)
\( \Rightarrow m = 0 \text{ or } 6 \)
In simple words: We use the distance formula with the given distance. After squaring and simplifying, we get a quadratic equation in m.

📝 Teacher's Note: Show students how to handle surds (like 3√5) by squaring both sides. Emphasize checking both solutions in the original equation.

🎯 Exam Tip: When distance involves surds, square both sides carefully. Simplify step by step. Always check if both solutions are valid for the given problem.

Answer 9. Find the relation between a and b if the points A(6, -1), B(5, 8) and P(a, b) are such that PA = PB.
Answer:

[Diagram: This shows a triangle with vertices A(6,-1), B(5,8), and P(a,b), where PA = PB, meaning P is equidistant from A and B.]

Given, PA = PB
∴ PA² = PB²
\( \Rightarrow (a - 6)^2 + (b + 1)^2 = (a - 5)^2 + (b - 8)^2 \)
\( \Rightarrow a^2 + 36 - 12a + b^2 + 1 + 2b = a^2 + 25 - 10a + b^2 + 64 - 16b \)
\( \Rightarrow -2a + 18b - 52 = 0 \)
\( \Rightarrow -a + 9b - 26 = 0 \)
\( \Rightarrow a = 9b - 26 \)
In simple words: When two distances are equal, we can square both sides to avoid square roots. This gives us a simple linear relationship between a and b.

📝 Teacher's Note: Explain that PA = PB means P lies on the perpendicular bisector of line segment AB. This creates a linear relationship between coordinates.

🎯 Exam Tip: When two distances are equal, square both sides immediately. Expand brackets carefully and collect like terms. The relation will be linear, not quadratic.

Answer 10. Find the locus of a point which is equidistant from the points R(0, 9) and T(14, 11).
Answer:

[Diagram: This shows two points R(0,9) and T(14,11) with point M(x,y) equidistant from both, forming a perpendicular bisector.]

Given: MR = MT
∴ MR² = MT²
\( (x - 0)^2 + (y - 9)^2 = (x - 14)^2 + (y - 11)^2 \)
\( x^2 + y^2 + 81 - 18y = x^2 + 196 - 28x + y^2 + 121 - 22y \)
\( 81 - 18y = 196 - 28x + 121 - 22y \)
\( 28x - 18y + 22y = 196 + 121 - 81 \)
\( 28x + 4y = 236 \)
\( 7x + y - 58 = 0 \)
In simple words: The locus is all points that are the same distance from R and T. This forms a straight line called the perpendicular bisector.

📝 Teacher's Note: Show students that the locus of equidistant points is always a straight line - the perpendicular bisector. Draw this on a graph for clarity.

🎯 Exam Tip: For locus problems, let the point be (x,y). Set up the distance equation and simplify. The locus equation will usually be linear for equidistant problems.

Answer 11. Find the distance between the points P and Q where P lies on the y-axis with ordinate 5 and Q lies on the x-axis with abscissa 12.
Answer: P lies on y-axis and has ordinate 5
∴ P(0, 5)
Q lies on x-axis and has an abscissa
∴ Q(12, 0)
\( ∴ PQ = \sqrt{(12 - 0)^2 + (0 - 5)^2} \)
\( = \sqrt{144 + 25} \)
\( = \sqrt{169} \)
\( = 13 \text{ units} \)
In simple words: P is on the y-axis so its x-coordinate is 0. Q is on the x-axis so its y-coordinate is 0. We use the distance formula to find PQ.

📝 Teacher's Note: Remind students that points on axes have one coordinate as zero. Y-axis means x = 0, X-axis means y = 0. Draw this on a graph.

🎯 Exam Tip: When a point lies on an axis, one coordinate is always zero. Remember: y-axis has x = 0, x-axis has y = 0. Use this to write coordinates correctly.

Answer 12.
Answer:
P lies on x-axis and Q lies on y-axis
Let abscissa of P be x then ordinate of Q is x-1.
∴ P(x, 0), Q(0, x - 1)
Given PQ = 5 units
\[ \sqrt{(x - 0)^2 + (0 - x + 1)^2} = 5 \]
squaring both sides
\[ x^2 + x^2 + 1 - 2x = 25 \]
\[ 2x^2 - 2x - 24 = 0 \]
\[ x^2 - x - 12 = 0 \]
\[ x^2 - 4x + 3x - 12 = 0 \]
\[ (x - 4)(x + 3) = 0 \]
x = +4 or -3
Coordinates of P are (4, 0) or (-3, 0)
Coordinates of Q are (0, 3) or (0, -4).
In simple words: We place P on x-axis and Q on y-axis. We use the distance formula to find where these points can be so their distance is 5 units.

📝 Teacher's Note: Draw the x-axis and y-axis on board. Show students how P has y-coordinate 0 and Q has x-coordinate 0. This makes the distance formula easier to understand.

🎯 Exam Tip: Always write the coordinates clearly at the end. Check your answer by putting the coordinates back in the distance formula.

Answer 13.
Answer:
Let the point on x-axis be P (x, 0).

[Diagram: Triangle with vertices A(5, 4), B(-2, 3), and point P(x, 0) on x-axis, showing PA = PB condition]

Given,
PA = PB
\[ PA^2 = PB^2 \]
\[ (x - 5)^2 + (0 - 4)^2 = (x + 2)^2 + (0 - 3)^2 \]
\[ x^2 + 25 - 10x + 16 = x^2 + 4 + 4x + 9 \]
\[ \Rightarrow -14x + 28 = 0 \]
\[ \Rightarrow 14x = 28 \]
\[ \Rightarrow x = 2 \]
∴ The point on x - axis is (2, 0)
In simple words: We find a point on x-axis that is equally far from both given points A and B. This point is at equal distance from both A and B.

📝 Teacher's Note: Tell students that when a point is equally far from two points, it lies on the perpendicular bisector. For x-axis points, y-coordinate is always 0.

🎯 Exam Tip: Write PA = PB first, then square both sides to remove the square root. This makes calculation easier. Always check your final answer.

Answer 14.
Answer:
A (-4, 3). Let the other point B (x, 9).
Given, AB = 10 units
\[ \sqrt{(-4 - x)^2 + (3 - 9)^2} = 10 \]
squaring both sides,
\[ \Rightarrow 16 + x^2 + 8x + 36 = 100 \]
\[ \Rightarrow x^2 + 8x - 48 = 0 \]
\[ \Rightarrow x^2 + 12x - 4x - 48 = 0 \]
\[ \Rightarrow x(x + 12) - 4(x + 12) = 0 \]
\[ \Rightarrow (x - 4)(x + 12) = 0 \]
\[ \Rightarrow x = 4 \text{ or } -12 \]
The abscissa of other end is 4 or -12.
In simple words: We know one end point and the distance between two points. We use the distance formula to find the x-coordinate of the other end point.

📝 Teacher's Note: Show students that one distance can give two possible points. Draw this on coordinate plane to make it clear why we get two answers.

🎯 Exam Tip: Write "abscissa" means x-coordinate. Give both possible values as the final answer. Don't forget to solve the quadratic equation completely.

Answer 15.
Answer:
A (5, 5), B (3, 4), C (-7, -1)
\[ AB = \sqrt{(5-3)^2 + (5-4)^2} = \sqrt{4+1} = \sqrt{5} \text{ units} \]
\[ BC = \sqrt{(3+7)^2 + (4+1)^2} = \sqrt{100+25} = 5\sqrt{5} \text{ units} \]
\[ AC = \sqrt{(5+7)^2 + (5+1)^2} = \sqrt{144+36} = 6\sqrt{5} \text{ units} \]
\[ AB + BC = \sqrt{5} + 5\sqrt{5} = 6\sqrt{5} = AC \]
∴ AB + BC = AC
∴ A, B and C are collinear points

P(5, 1), Q(3, 2), R(1, 3)
\[ PQ = \sqrt{(5-3)^2 + (1-2)^2} = \sqrt{4+1} = \sqrt{5} \text{ units} \]
\[ QR = \sqrt{(3-1)^2 + (2-3)^2} = \sqrt{4+1} = \sqrt{5} \text{ units} \]
\[ PR = \sqrt{(5-1)^2 + (1-3)^2} = \sqrt{16+4} = 2\sqrt{5} \text{ units} \]
\[ PQ + QR = \sqrt{5} + \sqrt{5} = 2\sqrt{5} = PR \]
∴ PQ + QR = PR
∴ P, Q and R are collinear points

M(4, -5), N(1, 1), S(-2, 7)
\[ MN = \sqrt{(4-1)^2 + (-5-1)^2} = \sqrt{9+36} = 3\sqrt{5} \text{ units} \]
\[ NS = \sqrt{(1+2)^2 + (1-7)^2} = \sqrt{9+36} = 3\sqrt{5} \text{ units} \]
\[ MS = \sqrt{(4+2)^2 + (-5-7)^2} = \sqrt{36+144} = 6\sqrt{5} \text{ units} \]
\[ MN + NS = 3\sqrt{5} + 3\sqrt{5} = 6\sqrt{5} = MS \]
∴ MN + NS = MS
∴ M, N and S are collinear points.
In simple words: When three points lie on the same straight line, the sum of two shorter distances equals the longest distance. We check this condition for all three sets of points.

📝 Teacher's Note: Draw three points on a line and show students that the middle point divides the line into two parts. The total distance is the sum of these two parts.

🎯 Exam Tip: Always calculate all three distances first. Then check if sum of two smaller distances equals the largest distance. Write "collinear points" clearly in your conclusion.

Answer 16.
Answer:

[Diagram: Circle with center O(x, y) and three points A(8, 12), B(11, 3), and C(0, 14) on the circumference]

Let O (x, y) be the centre of the circle.
OA = OB (radii of the same circle)
\[ \Rightarrow OA^2 = OB^2 \]
\[ (x-8)^2 + (y-12)^2 = (x-11)^2 + (y-3)^2 \]
\[ \Rightarrow x^2 + 64 - 16x + y^2 + 144 - 24y = x^2 + 121 - 22x + y^2 + 9 - 6y \]
\[ \Rightarrow 6x - 18y + 78 = 0 \]
\[ \Rightarrow x - 3y + 13 = 0 \] .....(1)
similarly, OB = OC
∴ OB² = OC²
\[ (x-11)^2 + (y-3)^2 = (x-0)^2 + (y-14)^2 \]
\[ \Rightarrow x^2 + 121 - 22x + y^2 + 9 - 6y = x^2 + y^2 + 196 - 28y \]
\[ \Rightarrow -22x + 22y - 66 = 0 \]
\[ \Rightarrow -x + y - 3 = 0 \] .....(2)
x - 3y + 13 = 0 .....(1)
solving (1) & (2) we get,
-2y + 10 = 0
\[ \Rightarrow y = 5 \]
from (1)
x - 15 + 13 = 0
\[ \Rightarrow x = 2 \]
Thus, coordinates of O are (2, 5)
\[ \text{Radius} = \sqrt{(2-8)^2 + (5-12)^2} = \sqrt{36+49} = \sqrt{85} \text{ units} \]
In simple words: The centre of a circle is equally far from all points on the circle. We use this fact to find equations and solve them to get the centre coordinates.

📝 Teacher's Note: Explain that all radii of a circle are equal. Use this property to form equations. Show how solving two equations gives the centre coordinates.

🎯 Exam Tip: Write "radii are equal" clearly. Form two equations using this fact. Solve the system of equations step by step. Calculate radius using any one point and the centre.

Answer 19.
Answer:

[Diagram: Circle with center O and point P(8, 5) outside the circle, with diameter = 20 units]

Given diameter of the circle = 20 units.
∴ radius = 10 units
OP = 10
\[ \sqrt{(x+1-8)^2 + (x-4-5)^2} = 10 \]
squaring both sides,
\[ x^2 + 49 - 14x + x^2 = 81 - 18x = 100 \]
\[ \Rightarrow 2x^2 - 32x + 30 = 0 \]
\[ \Rightarrow x^2 - 16x + 15 = 0 \]
\[ \Rightarrow x^2 - 15x - x + 15 = 0 \]
\[ \Rightarrow (x-15)(x-1) = 0 \]
\[ \Rightarrow x = 15 \text{ or } 1 \]
Coordinates of O when x = 15 are (16, 11)
Coordinates of O when x = 1 are (2, -3)
In simple words: The centre of the circle is at distance equal to radius from point P. We use this condition to find two possible positions of the centre.

📝 Teacher's Note: Draw a circle and show that there can be two circles of same radius passing through a given external point. This explains why we get two answers.

🎯 Exam Tip: Write radius = diameter ÷ 2 first. Use the condition that centre is at distance equal to radius from the given point. Give both possible coordinates of the centre.

Answer 23.
Answer:

[Diagram: Triangle with vertices A(6, -1), B(5, 8), and C(1, 3)]

A (6, -1), B (5, 8), C (1, 3)
\[ AB = \sqrt{(6-5)^2 + (-1-8)^2} = \sqrt{1+81} = \sqrt{82} \text{ units} \]
\[ BC = \sqrt{(5-1)^2 + (8-3)^2} = \sqrt{16+25} = \sqrt{41} \text{ units} \]
\[ AC = \sqrt{(1-6)^2 + (3-1)^2} = \sqrt{25+16} = \sqrt{41} \text{ units} \]
∴ BC = AC,
∴ A, B and C are the vertices of an isosceles triangle.
In simple words: An isosceles triangle has two sides of equal length. We calculate all three sides and find that BC = AC, so this is an isosceles triangle.

📝 Teacher's Note: Remind students that isosceles means two equal sides. Draw the triangle and mark the two equal sides clearly with same symbols.

🎯 Exam Tip: Calculate all three side lengths using distance formula. Compare the lengths and identify which two are equal. State clearly that it is an isosceles triangle.

Answer 24.
Answer:

[Diagram: Triangle with vertices P(1, 1), Q(-4, 4), and R(4, 6)]

P (1, 1), Q (-4, 4), R (4, 6)
\[ PQ = \sqrt{(1+4)^2 + (1-4)^2} = \sqrt{25+9} = \sqrt{34} \text{ units} \]
\[ QR = \sqrt{(-4-4)^2 + (4-6)^2} = \sqrt{64+4} = \sqrt{68} \text{ units} \]
\[ PR = \sqrt{(4-1)^2 + (6-1)^2} = \sqrt{9+25} = \sqrt{34} \text{ units} \]
∴ PQ = QR,
∴ P, Q and R are the vertices of an isosceles triangle
In simple words: We calculate the three side lengths and find that PQ = PR. Since two sides are equal, this triangle is isosceles.

📝 Teacher's Note: Point out that the vertex with equal sides (P) is called the apex of the isosceles triangle. The side opposite to it (QR) is called the base.

🎯 Exam Tip: Use distance formula correctly for all three sides. Compare the results and identify the two equal sides. Write the conclusion about triangle type clearly.

Answer 25.
Answer:

[Diagram: Triangle with vertices A(-2, 1), B(0, 3), and C(-1, 4)]

A (-2, 1), B (0, 3), C (-1, 4)
\[ AB = \sqrt{(-2-0)^2 + (1-3)^2} = \sqrt{4+4} = \sqrt{8} \text{ units} \]
\[ BC = \sqrt{(0+1)^2 + (3-4)^2} = \sqrt{1+1} = \sqrt{2} \text{ units} \]
\[ AC = \sqrt{(-2+1)^2 + (1-4)^2} = \sqrt{1+9} = \sqrt{10} \text{ units} \]
\[ AB^2 + BC^2 = 8 + 2 = 10 \]
\[ AC^2 = 10 \]
∴ \( AB^2 + BC^2 = AC^2 \)
∴ A, B and C are the vertices of a right angled triangle.
In simple words: We use Pythagoras theorem to check if this is a right triangle. Since the square of the longest side equals the sum of squares of the other two sides, it is a right triangle.

📝 Teacher's Note: Explain Pythagoras theorem: \( a^2 + b^2 = c^2 \) where c is the longest side. Show students how to identify the longest side first.

🎯 Exam Tip: Always find the longest side first (biggest value). Check if its square equals sum of squares of other two sides. Write the Pythagoras condition clearly to get full marks.

Answer 26.
Answer:
Given triangle with vertices A(7, 10), B(-2, 5), and C(3, -4).

Step 1: Find the length of side AB.
\[ AB = \sqrt{(7 + 2)^2 + (10 - 5)^2} = \sqrt{81 + 25} = \sqrt{106} \text{ units} \]

Step 2: Find the length of side BC.
\[ BC = \sqrt{(-2 - 3)^2 + (5 + 4)^2} = \sqrt{25 + 81} = \sqrt{106} \text{ units} \]

Step 3: Find the length of side AC.
\[ AC = \sqrt{(7 - 3)^2 + (10 + 4)^2} = \sqrt{16 + 196} = \sqrt{212} \text{ units} \]

Step 4: Check if triangle is isosceles.
Since AB = BC, the triangle ABC is an isosceles triangle.

Step 5: Check if triangle is right-angled using Pythagoras theorem.
\[ AB^2 + BC^2 = 100 + 106 = 212 \]
\[ AC^2 = 212 \]
Since \( AB^2 + BC^2 = AC^2 \), triangle ABC is also a right-angled triangle.

In simple words: We found all three side lengths using distance formula. Two sides are equal, so it is isosceles. Also, the square of the longest side equals sum of squares of other two sides, so it is right-angled.

📝 Teacher's Note: Draw the triangle on graph paper to show students visually. Mark equal sides with same symbols. Show how Pythagoras theorem works with actual numbers.

🎯 Exam Tip: Always check both conditions separately. Write "AB = BC, so isosceles" and "AB² + BC² = AC², so right-angled". Show all calculations clearly.

Answer 27.
Answer:
Given triangle with vertices P(1, 1), Q(-√3, √3), and R(-1, -1).

Step 1: Find the length of side PQ.
\[ PQ = \sqrt{(1 + \sqrt{3})^2 + (1 - \sqrt{3})^2} = \sqrt{4 + 4} = \sqrt{8} \text{ units} \]

Step 2: Find the length of side QR.
\[ QR = \sqrt{(-\sqrt{3} + 1)^2 + (\sqrt{3} + 1)^2} = \sqrt{4 + 4} = \sqrt{8} \text{ units} \]

Step 3: Find the length of side PR.
\[ PR = \sqrt{(-1 - 1)^2 + (-1 - 1)^2} = \sqrt{4 + 4} = \sqrt{8} \text{ units} \]

Step 4: Identify the type of triangle.
Since PQ = QR = PR, triangle PQR is an equilateral triangle.

In simple words: All three sides have the same length. When all sides are equal, the triangle is called equilateral. All angles are also 60° each.

📝 Teacher's Note: Explain that equilateral means "equal sides". All angles in equilateral triangle are 60°. Students should remember this special property.

🎯 Exam Tip: Write "PQ = QR = PR = √8 units, so triangle is equilateral". Always mention that all sides are equal and give the measurement.

Answer 28.
Answer:
Given triangle with vertices A(0, 3), B(4, 3), and C(2, 3 + 2√5).

Step 1: Find the length of side AB.
\[ AB = \sqrt{(0 - 4)^2 + (3 - 3)^2} = 4 \text{ units} \]

Step 2: Find the length of side BC.
\[ BC = \sqrt{(4 - 2)^2 + (3 - 3 - 2\sqrt{5})^2} = \sqrt{4 + 12} = 4 \text{ units} \]

Step 3: Find the length of side AC.
\[ AC = \sqrt{(2 - 0)^2 + (3 + 2\sqrt{3} - 3)^2} = \sqrt{4 + 12} = 4 \text{ units} \]

Step 4: Identify the type of triangle.
Since AB = BC = AC, triangle ABC is an equilateral triangle.

In simple words: All three sides are 4 units long. When all sides are equal, it is an equilateral triangle. This means all angles are also equal at 60° each.

📝 Teacher's Note: Show students that even with complex coordinates involving square roots, the final answer can be simple. Emphasize careful calculation of distance formula.

🎯 Exam Tip: Double-check your distance calculations, especially with square roots. Write "All sides = 4 units, so equilateral triangle" as your conclusion.

Answer 29.
Answer:
Given quadrilateral with vertices A(5, 3), B(1, 2), C(2, -2), and D(6, -1).

Step 1: Find the length of all sides.
\[ AB = \sqrt{(5 - 1)^2 + (3 - 2)^2} = \sqrt{16 + 1} = \sqrt{17} \text{ units} \]
\[ BC = \sqrt{(1 - 2)^2 + (2 + 2)^2} = \sqrt{1 + 16} = \sqrt{17} \text{ units} \]
\[ CD = \sqrt{(6 - 2)^2 + (-1 + 2)^2} = \sqrt{16 + 1} = \sqrt{17} \text{ units} \]
\[ DA = \sqrt{(6 - 5)^2 + (-1 - 3)^2} = \sqrt{1 + 16} = \sqrt{17} \text{ units} \]

Step 2: Find the length of diagonals.
\[ AC = \sqrt{(5 - 2)^2 + (3 + 2)^2} = \sqrt{9 + 25} = \sqrt{34} \text{ units} \]
\[ BD = \sqrt{(6 - 1)^2 + (-1 - 2)^2} = \sqrt{25 + 9} = \sqrt{34} \text{ units} \]

Step 3: Identify the type of quadrilateral.
Since AB = BC = CD = DA and AC = BD, quadrilateral ABCD is a square.

In simple words: All four sides are equal and both diagonals are equal. This makes it a square. A square is a special rectangle where all sides are the same length.

📝 Teacher's Note: Remind students that a square has equal sides AND equal diagonals. A rhombus has equal sides but unequal diagonals. This distinction is important.

🎯 Exam Tip: Always check both sides and diagonals for quadrilaterals. Write "All sides equal AND diagonals equal, so it is a square" for full marks.

Answer 30.
Answer:
Given quadrilateral with vertices A(4, 6), B(-1, 5), C(-2, 0), and D(3, 1).

Step 1: Find the length of all sides.
\[ AB = \sqrt{(4 + 1)^2 + (6 - 5)^2} = \sqrt{25 + 1} = \sqrt{26} \text{ units} \]
\[ BC = \sqrt{(-1 + 2)^2 + (5 - 0)^2} = \sqrt{1 + 25} = \sqrt{26} \text{ units} \]
\[ CD = \sqrt{(-2 - 3)^2 + (0 - 1)^2} = \sqrt{25 + 1} = \sqrt{26} \text{ units} \]
\[ DA = \sqrt{(3 - 4)^2 + (1 - 6)^2} = \sqrt{1 + 25} = \sqrt{26} \text{ units} \]

Step 2: Find the length of diagonals.
\[ AC = \sqrt{(4 + 2)^2 + (6 - 0)^2} = \sqrt{36 + 36} = 6\sqrt{2} \text{ units} \]
\[ BD = \sqrt{(-1 - 3)^2 + (5 - 1)^2} = \sqrt{36 + 36} = 16\sqrt{2} \text{ units} \]

Step 3: Identify the type of quadrilateral.
Since AB = BC = CD = DA and AC ≠ BD, quadrilateral ABCD is a rhombus.

In simple words: All four sides are equal but the diagonals are not equal. This makes it a rhombus. A rhombus is like a square that has been pushed sideways.

📝 Teacher's Note: Draw a square and a rhombus side by side. Show how rhombus has equal sides but different diagonal lengths. This visual helps students remember the difference.

🎯 Exam Tip: For rhombus, write "All sides equal but diagonals unequal". This key phrase shows you know the defining properties of a rhombus.

Answer 31.
Answer:
Given quadrilateral with vertices A(0, 0), B(3, 2), C(7, 7), and D(4, 5).

Step 1: Find the length of opposite sides.
\[ AB = \sqrt{(3 - 0)^2 + (2 - 0)^2} = \sqrt{9 + 4} = \sqrt{13} \text{ units} \]
\[ BC = \sqrt{(3 - 7)^2 + (2 - 7)^2} = \sqrt{16 + 25} = \sqrt{41} \text{ units} \]
\[ CD = \sqrt{(7 - 4)^2 + (7 - 5)^2} = \sqrt{9 + 4} = \sqrt{13} \text{ units} \]
\[ DA = \sqrt{(4 - 0)^2 + (5 - 0)^2} = \sqrt{16 + 25} = \sqrt{41} \text{ units} \]

Step 2: Identify the type of quadrilateral.
Since AB = CD and BC = DA, quadrilateral ABCD is a parallelogram.

In simple words: Opposite sides are equal in length. When opposite sides are equal, the shape is a parallelogram. Think of it like a rectangle that has been tilted.

📝 Teacher's Note: Show students that in parallelogram, opposite sides are parallel AND equal. Use a simple drawing to show how opposite sides never meet (parallel).

🎯 Exam Tip: Write "AB = CD and BC = DA, so opposite sides are equal. Therefore, ABCD is a parallelogram." This shows clear reasoning.

Answer 32.
Answer:
Given quadrilateral with vertices A(0, 2), B(1, 1), C(4, 4), and D(3, 5).

Step 1: Find the length of all sides.
\[ AB = \sqrt{(0 - 1)^2 + (2 - 1)^2} = \sqrt{2} \text{ units} \]
\[ BC = \sqrt{(1 - 4)^2 + (1 - 4)^2} = 3\sqrt{2} \text{ units} \]
\[ CD = \sqrt{(4 - 3)^2 + (4 - 5)^2} = \sqrt{2} \text{ units} \]
\[ DA = \sqrt{(3 - 0)^2 + (5 - 2)^2} = 3\sqrt{2} \text{ units} \]

Step 2: Find the length of diagonals.
\[ AC = \sqrt{(4 - 0)^2 + (4 - 2)^2} = \sqrt{20} = 2\sqrt{5} \text{ units} \]
\[ BC = \sqrt{(3 - 1)^2 + (5 - 1)^2} = \sqrt{20} = 2\sqrt{5} \text{ units} \]

Step 3: Identify the type of quadrilateral.
Since AB = CD and BC = DA, and also AC = BD, quadrilateral ABCD is a rectangle.

In simple words: Opposite sides are equal and diagonals are equal. This makes it a rectangle. A rectangle has four right angles and opposite sides are parallel and equal.

📝 Teacher's Note: Explain that rectangle is a special parallelogram where all angles are 90°. The equal diagonals make it different from a general parallelogram.

🎯 Exam Tip: Write "Opposite sides equal AND diagonals equal, so it is a rectangle." The key is mentioning both conditions for rectangle.

Answer 33.
Answer:
Given quadrilateral PQRS with vertices P(a, b), Q(a+3, b+4), R(a-1, b+7), and S(a-4, b+3).

Step 1: Find the length of opposite sides.
\[ PQ = \sqrt{(a + 3 - a)^2 + (b + 4 - b)^2} = \sqrt{9 + 16} = 5 \text{ units} \]
\[ QR = \sqrt{(a + 3 - a + 1)^2 + (b + 4 - b - 7)^2} = \sqrt{16 + 9} = 5 \text{ units} \]
\[ RS = \sqrt{(a - 1 - a + 4)^2 + (b + 7 - b - 3)^2} = \sqrt{9 + 16} = 5 \text{ units} \]
\[ SP = \sqrt{(a - 4 - a)^2 + (b + 3 - b)^2} = \sqrt{16 + 9} = 5 \text{ units} \]

Step 2: Analyze the result.
Since the opposite sides of quadrilateral PQRS are equal, therefore, it is a parallelogram.

In simple words: Even though the coordinates have variables (a and b), when we calculate distances, all sides come out equal. This makes it a special parallelogram where all sides are equal.

📝 Teacher's Note: This problem shows how algebra works with geometry. The variables (a, b) represent any point, but the relative positions make it a parallelogram always.

🎯 Exam Tip: When coordinates have variables, calculate distances carefully. Focus on the differences between coordinates. Write your final conclusion clearly about the shape type.

Ex 12.2

Answer 1.
Answer:
(a)
Let the point P divide the line segment AB in the ratio 1:2.
∴ coordinates of P are
\( x = \frac{1 \times 6 + 2 \times 3}{1 + 2} = 4 \)
\( y = \frac{1 \times 9 + 2 \times (-3)}{1 + 2} = 1 \)
P(4, 1)

(b)
Let the point P divide the line segment MN in the ratio 2:5.
∴ coordinates of P are
\( x = \frac{2 \times 3 + 5 \times (-4)}{2 + 5} = \frac{-14}{7} = -2 \)
\( y = \frac{2 \times 2 + 5 \times (-5)}{2 + 5} = -3 \)
P(-2, -3)

(c)
Let the point P divide the line segment SR in the ratio 3:4.
∴ coordinates of P are
\( x = \frac{3 \times 9 + 4 \times 2}{3 + 4} = 5 \)
\( y = \frac{3 \times (-8) + 4 \times 6}{3 + 4} = 0 \)
P(5, 0)

(d)
Let the point P divide DE in the ratio 4:7.
∴ coordinates of P are
\( x = \frac{4 \times 15 + 7 \times (-7)}{4 + 7} = 1 \)
\( y = \frac{4 \times (-2) + 7 \times 9}{4 + 7} = 5 \)
P(1, 5)

In simple words: This uses the section formula to find where a point divides a line. We use the given ratio to find the new coordinates.

📝 Teacher's Note: Draw the line segments on graph paper. Show students how the section formula works like finding a weighted average. The ratio tells us how to split the line.

🎯 Exam Tip: Always write the section formula first. Substitute values carefully. Check your arithmetic twice to avoid mistakes.

Answer 2.
Answer:
Let P(x, y) and Q(a, b) be the point of trisection of the line segment AB.
AP : PB = 1:2
Coordinates of P are
\( x = \frac{1 \times 3 + 2 \times (-3)}{1 + 2} = -1 \)
\( y = \frac{1 \times (-2) + 2 \times 7}{1 + 2} = 4 \)
P(-1, 4)

AQ : QB = 2:1
coordinates of Q are,
\( a = \frac{2 \times 3 + 1 \times (-3)}{2 + 1} = 1 \)
\( b = \frac{2 \times (-2) + 1 \times 7}{2 + 1} = 1 \)
Q(1, 1)

∴ The points of trisection are (-1, 4) and (1, 1).

In simple words: Trisection means cutting the line into 3 equal parts. We get 2 points that divide the line equally.

📝 Teacher's Note: Explain that trisection means three equal parts. One point divides in 1:2 ratio, the other in 2:1 ratio. Draw this on the board.

🎯 Exam Tip: For trisection, always use ratios 1:2 and 2:1. Write both points clearly. Label them properly as P and Q.

Answer 3.
Answer:
Let A(x, y) and B(a, b) be the points of trisection of line segment MN.
MA : AN = 1 : 2
∴ coordinates of A are,
\( x = \frac{1 \times 6 + 2 \times 3}{1 + 2} = 4 \)
\( y = \frac{1 \times 9 + 2 \times (-3)}{1 + 2} = 1 \)
A(4, 1)

Also, MB : BN = 2 : 1
coordinates of B are,
\( a = \frac{2 \times 6 + 1 \times 3}{2 + 1} = 5 \)
\( b = \frac{2 \times 9 + 1 \times (-3)}{2 + 1} = 5 \)
B(5, 5)

points of trisection are (4, 1) and (5, 5).

In simple words: We divide line MN into three equal parts. This gives us two dividing points with coordinates (4, 1) and (5, 5).

📝 Teacher's Note: Show students that the two trisection points are different from the previous problem because the endpoints M and N are different.

🎯 Exam Tip: Always identify which point is which (first trisection point vs second). Use the correct ratios 1:2 and 2:1.

Answer 4.
Answer:
Let the point P divide AB in the ratio k:1.
Coordinates of P are,
\( x = \frac{7k - 3}{k + 1} \)
\( y = \frac{6k + 1}{k + 1} \)

But given, P(x, y) = P(2, 4)

∴ \( 2 = \frac{7k - 3}{k + 1} \)
\( \Rightarrow 2k + 2 = 7k - 3 \)
\( \Rightarrow 5 = 5k \)
\( \Rightarrow k = 1 \)
k : 1 = 1 : 1

or \( 4 = \frac{6k + 1}{k + 1} \)
\( 4k + 4 = 6k + 1 \)
\( \Rightarrow 3 = 2k \)
\( \Rightarrow k = \frac{3}{2} \)
k : 1 = 3 : 2

In simple words: We find what ratio k:1 gives us the point P(2, 4). We can check using either x-coordinate or y-coordinate.

📝 Teacher's Note: Show students they can use either x or y coordinate to find k. Both should give the same answer if calculations are correct.

🎯 Exam Tip: Set up the section formula equation using the given point. Solve for k step by step. Check your answer using the other coordinate.

Answer 5.
Answer:
Let R divide the line segment ST in the ratio k : 1.
Coordinates of R
R(x, y) = R(1, 5)
\( R\left(\frac{5k - 2}{k + 1}, \frac{13k - 1}{k + 1}\right) = R(1, 5) \)

\( \frac{5k - 2}{k + 1} = 1 \)
5k - 2 = k + 1
4k = 3
\( k = \frac{3}{4} \)

Hence, required ratio is k : 1 = 3 : 4.

In simple words: Point R(1, 5) divides line ST in the ratio 3:4. We found this by using the section formula and solving for k.

📝 Teacher's Note: Remind students that the ratio can be written as 3:4, which means R is closer to S than to T on the line segment.

🎯 Exam Tip: Always write the final ratio in simplest form. Convert fractions like 3/4 to the ratio 3:4 clearly.

Answer 34.
Answer:
[Diagram: A rectangle ABCD with vertices A(0, -4), B(6, 2), C(3, 5), and D(-3, -1)]

\( AB = \sqrt{(6 - 0)^2 + (2 + 4)^2} = 6\sqrt{2} \text{ units} \)
\( BC = \sqrt{(6 - 3)^2 + (2 - 5)^2} = 3\sqrt{2} \text{ units} \)
\( CD = \sqrt{(3 + 3)^2 + (5 + 1)^2} = 6\sqrt{2} \text{ units} \)
\( DA = \sqrt{(-3 - 0)^2 + (-1 + 4)^2} = 3\sqrt{2} \text{ units} \)
\( AC = \sqrt{(3 - 0)^2 + (5 + 4)^2} = 3\sqrt{10} \text{ units} \)
\( BD = \sqrt{(6 + 3)^2 + (2 + 1)^2} = 3\sqrt{10} \text{ units} \)

∴ AB = CD and BC = DA,
Also AC = BD
∴ ABCD is a rectangle.

In simple words: We found the lengths of all sides and diagonals. Opposite sides are equal and diagonals are equal, so ABCD is a rectangle.

📝 Teacher's Note: Show students that for a rectangle, opposite sides are equal and diagonals are equal. Use the distance formula to calculate each length.

🎯 Exam Tip: Calculate all four sides and both diagonals. Write clearly that opposite sides are equal and diagonals are equal to prove it's a rectangle.

Answer 37.
Answer:
[Diagram: Triangle ABC with A(1, 1), B(-1, -1), and C(x, y) where ABC is equilateral]

ABC is an equilateral triangle.
∴ AC = BC and AB = BC
\( \Rightarrow AC^2 = BC^2 \) and \( AB^2 = BC^2 \)

\( (x - 1)^2 + (y - 1)^2 = (x + 1)^2 + (y + 1)^2 \)
\( \Rightarrow x^2 + 1 - 2x + y^2 + 1 - 2y = x^2 + 1 + 2x + y^2 + 1 + 2y \)
\( \Rightarrow -4x - 4y = 0 \)
\( \Rightarrow -4x = 4y \)
\( \Rightarrow x = -y \) ........(1)

\( (1 + 1)^2 + (1 + 1)^2 = (x + 1)^2 + (y + 1)^2 \)
\( \Rightarrow 8 = x^2 + 1 + 2x + y^2 + 1 + 2y \)
\( \Rightarrow 8 = y^2 + 1 - 2y + y^2 + 1 + 2y \)
\( \Rightarrow 2y^2 - 6 = 0 \)
\( \Rightarrow y^2 = 3 \)
\( \Rightarrow y = \pm \sqrt{3} \)

from (1)
∴ \( x = \mp \sqrt{3} \)

In simple words: For an equilateral triangle, all three sides must be equal. Using this condition, we find that C can be at \( (\sqrt{3}, -\sqrt{3}) \) or \( (-\sqrt{3}, \sqrt{3}) \).

📝 Teacher's Note: Explain that equilateral means all sides equal. Set up equations using the distance formula and solve step by step. There are two possible positions for C.

🎯 Exam Tip: Write "equilateral triangle means all sides equal" first. Use \( AC^2 = BC^2 \) and \( AB^2 = BC^2 \) to set up equations. Find both possible answers.

Answer 6.
Answer: Given that points A(x, y), B(a, b) and C(p, q) divide the line segment MN in four equal parts, where M(-1, 10) and N(7, -2).

Step 1: Find coordinates of B (midpoint of MN).
\( B(a,b) = B\left(\frac{7-1}{2}, \frac{-2+10}{2}\right) = B(3,4) \)

Step 2: Find coordinates of A (midpoint of MB).
Since MA : AB = 1 : 1
\( A(x,y) = A\left(\frac{3-1}{2}, \frac{4+10}{2}\right) = A(1,7) \)

Step 3: Find coordinates of C (midpoint of BN).
Since BC : CN = 1 : 1
\( C(p,q) = C\left(\frac{3+7}{2}, \frac{4-2}{2}\right) = C(5,1) \)

Hence, the coordinates of A, B and C are (1,7), (3,4) and (5,1) respectively.
In simple words: We find the middle points step by step. First the middle of the whole line, then the middle of each half.

📝 Teacher's Note: Draw a line segment and mark four equal parts. Show students how to find the middle point first, then the quarter points. This makes the pattern clear.

🎯 Exam Tip: Always find the midpoint of the whole segment first. Then find midpoints of the smaller segments. Write coordinates clearly in brackets.

Answer 7.
Answer: Let point P(x, 0) on x-axis divide line segment AB in the ratio k : 1, where A(-2, -3) and B(5, 6).

Step 1: Use section formula for x-coordinate.
\( x = \frac{5k + 2}{k + 1} \)

Step 2: Use section formula for y-coordinate.
\( 0 = \frac{6k - 3}{k + 1} \)

Step 3: Solve for k.
\( \Rightarrow 0 = 6k - 3 \)
\( \Rightarrow k = \frac{1}{2} \)

Hence, the required ratio is 1 : 2.
In simple words: Since the point is on x-axis, its y-coordinate is 0. We use this fact to find the ratio.

📝 Teacher's Note: Remind students that any point on x-axis has y-coordinate = 0. Use this condition to solve for the ratio.

🎯 Exam Tip: For points on x-axis, set y = 0 in section formula. For points on y-axis, set x = 0. This gives you the equation to solve.

Answer 8.
Answer: Given PQ is divided by line Y = 0 (x-axis). Let S(x, 0) be the point on line Y = 0 which divides line segment PQ in ratio k : 1, where P(4, -6) and Q(-3, 8).

Step 1: Use section formula.
\( x = \frac{-3k + 4}{k + 1}, \quad 0 = \frac{8k - 6}{k + 1} \)

Step 2: Solve for k using y-coordinate condition.
\( \Rightarrow 8k = 6 \)
\( \Rightarrow k = \frac{3}{4} \)

Hence, the required ratio is 3 : 4.
In simple words: We find where the line PQ crosses the x-axis. This gives us the ratio in which it is divided.

📝 Teacher's Note: Show students how any line crossing x-axis will have y = 0 at that crossing point. Use this to find the ratio.

🎯 Exam Tip: When a line segment is divided by an axis, the coordinate of that axis becomes zero. Use this to find k, then state the ratio clearly.

Answer 9.
Answer: Let point P(0, y) on y-axis divide line segment AB in ratio k : 1, where A(2, -1) and B(-5, 6).

Step 1: Use section formula.
\( 0 = \frac{-5k + 2}{k + 1}, \quad y = \frac{6k - 1}{k + 1} \)

Step 2: Solve for k using x-coordinate condition.
\( \Rightarrow 5k = 2 \)
\( \Rightarrow k = \frac{2}{5} \)

Hence, the required ratio is 2 : 5.
In simple words: We find where line AB crosses the y-axis. Since it's on y-axis, x-coordinate is 0.

📝 Teacher's Note: For y-axis problems, x = 0. Use this condition in section formula to find the ratio. Students often forget this simple rule.

🎯 Exam Tip: Points on y-axis have x = 0. Use this in section formula to find k. Then write the ratio in simplest form.

Answer 10.
Answer: Let P(x, 0) be the point on line y = 0 (x-axis) which divides line segment AB in ratio k : 1, where A(2, -4) and B(-3, 6).

Step 1: Use section formula.
\( x = \frac{-3k + 2}{k + 1}, \quad 0 = \frac{6k - 4}{k + 1} \)

Step 2: Solve for k.
\( \Rightarrow 6k = 4 \)
\( \Rightarrow k = \frac{2}{3} \)

Hence the required ratio is 2 : 3.
In simple words: We find the point where line AB meets the x-axis. At this point, y = 0.

📝 Teacher's Note: Draw the line segment AB and show where it crosses x-axis. The y-coordinate at crossing point is always zero.

🎯 Exam Tip: For x-axis intersection, y = 0. Use this condition to solve for k. Always simplify the ratio to lowest terms.

Answer 11.
Answer: Let S(0, y) be the point on line x = 0 (y-axis) which divides line segment PQ in ratio k : 1, where P(-4, 7) and Q(3, 0).

Step 1: Use section formula.
\( 0 = \frac{3k - 4}{k + 1}, \quad Y = \frac{0 + 7}{k + 1} \)

Step 2: Solve for k.
\( \Rightarrow 3k = 4 \)
\( k = \frac{4}{3} = \frac{4}{3} \)

Step 3: Find y-coordinate.
\( Y = \frac{7}{\frac{4}{3} + 1} = \frac{7}{\frac{7}{3}} = 3 \)

Hence, the required ratio is 4 : 3 and the required point is S(0, 3).
In simple words: We find where line PQ crosses the y-axis. At this crossing point, x = 0.

📝 Teacher's Note: Show students the line PQ on a graph. Mark where it crosses y-axis. This visual helps them understand the concept better.

🎯 Exam Tip: For y-axis problems, x = 0. Find k first, then substitute back to get the y-coordinate. Write both ratio and point coordinates.

Answer 12.
Answer: Let point P(1, a) divide line segment AB in ratio k : 1, where A(-1, 4) and B(4, -1).

Step 1: Use section formula for x-coordinate.
\( 1 = \frac{4k - 1}{k + 1} \)

Step 2: Solve for k.
\( \Rightarrow k + 1 = 4k - 1 \)
\( \Rightarrow 2 = 3k \)
\( \Rightarrow k = \frac{2}{3} \)

Step 3: Find value of a using y-coordinate.
\( a = \frac{-k + 4}{k + 1} = \frac{-\frac{2}{3} + 4}{\frac{2}{3} + 1} = \frac{\frac{10}{3}}{\frac{5}{3}} = 2 \)

Hence, the required ratio is 2 : 3 and the value of a is 2.
In simple words: We know the point P has x = 1. We use this to find k, then use k to find a.

📝 Teacher's Note: When one coordinate is given, use it to find k first. Then use that k to find the unknown coordinate. Do this step by step.

🎯 Exam Tip: Given partial coordinates, use the known coordinate in section formula to find k. Then substitute k back to find unknown coordinate.

Answer 13.
Answer: Given: -PB : AB = 1 : 5, so PB : PA = 1 : 4. Point P(x, y) divides AB externally in ratio 4 : 1, where A(-3, -10) and B(3, 2).

Step 1: Use external division formula.
\( (x,y) = \left(\frac{4 \times 3 - 3}{5}, \frac{4 \times 2 - 10}{5}\right) = \left(\frac{9}{5}, \frac{-2}{5}\right) \)

\( P\left(\frac{9}{5}, \frac{-2}{5}\right) \)
In simple words: External division means the point is outside the line segment, not between the two end points.

📝 Teacher's Note: External division formula is different from internal division. Make sure students understand the difference. Draw diagrams to show internal vs external division.

🎯 Exam Tip: For external division, use the external division formula. Be careful with signs - it's different from internal division formula.

Answer 14.
Answer: Let P(-2, y) be the point which divides line segment AB in ratio k : 1, where A(-6, -1) and B(1, 6).

Step 1: Use section formula for x-coordinate.
\( -2 = \frac{k - 6}{k + 1} \)

Step 2: Solve for k.
\( \Rightarrow -2k - 2 = k - 6 \)
\( \Rightarrow -3k = -4 \)
\( \Rightarrow k = \frac{4}{3} \)

Step 3: Find y-coordinate.
\( y = \frac{6k - 1}{k + 1} = \frac{6 \times \frac{4}{3} - 1}{\frac{4}{3} + 1} = \frac{8 - 1}{\frac{7}{3}} = \frac{21}{7} = 3 \)

Hence, the required ratio is 4 : 3 and the point of intersection is (-2, 3).
In simple words: We use the known x-coordinate to find the ratio, then find the y-coordinate using that ratio.

📝 Teacher's Note: When x-coordinate is given as -2, substitute this in section formula to find k. Then use k to find y-coordinate.

🎯 Exam Tip: Use the given coordinate to find k first. Then substitute k back to find the unknown coordinate. Show all working steps clearly.

Answer 15.
Answer: Let R(x, -1) be the point on line y = -1 which divides line segment PQ in ratio k : 1, where P(6, 5) and Q(-2, -11).

Step 1: Use section formula for y-coordinate.
\( -1 = \frac{-11k + 5}{k + 1} \)

Step 2: Solve for k.
\( \Rightarrow -k - 1 = -11k + 5 \)
\( \Rightarrow 10k = 6 \)
\( \Rightarrow k = \frac{3}{5} \)

Step 3: Find x-coordinate.
\( x = \frac{-2k + 6}{k + 1} = \frac{-2 \times \frac{3}{5} + 6}{\frac{3}{5} + 1} = \frac{-\frac{6}{5} + 6}{\frac{8}{5}} = \frac{\frac{24}{5}}{\frac{8}{5}} = 3 \)

Hence, the required ratio is 3 : 5 and the point of intersection is (3, -1).
In simple words: We use the fact that y = -1 to find the ratio, then find where this point lies on the x-axis.

📝 Teacher's Note: When a point lies on a horizontal line like y = -1, use this y-value in section formula to find the ratio first.

🎯 Exam Tip: For lines parallel to x-axis (y = constant), use the y-coordinate condition to find k. Then find x-coordinate.

Answer 16.
Answer: R(0, y) is the point on y-axis that divides PQ. Let the ratio in which PQ is divided by R be m : n. Given P(-5, 6) and Q(3, 2).
 

[Diagram: A coordinate plane showing points P(-5,6), Q(3,2), and R(0,y) on the y-axis, with lines connecting them.]

Step 1: Use section formula for x-coordinate.
\( 0 = \frac{mx_2 + nx_1}{m + n} = \frac{3m - 5n}{m + n} \)

Step 2: Solve the equation.
\( \Rightarrow 0 = 3m - 5n \)
\( \Rightarrow 3m = 5n \)
\( \Rightarrow \frac{m}{n} = \frac{5}{3} \)
\( \Rightarrow m : n = 5 : 3 \)
\( \Rightarrow PR : RQ = 5 : 3 \)
In simple words: Since R is on y-axis, its x-coordinate is 0. We use this to find the ratio.

📝 Teacher's Note: Draw the diagram clearly. Show students that when a point divides a segment and lies on y-axis, x = 0 gives the condition.

🎯 Exam Tip: For points on y-axis, x = 0. Set up the equation and solve for the ratio. Write the final answer as PR : RQ.

Exercise 12.3

Answer 1.
Answer:
(a)
[Diagram: Line segment AB with points A(4,7), P(x,y), and B(10,15) marked, showing midpoint calculation]
Coordinates of P are
\( P(x,y) = P\left(\frac{4 + 10}{2}, \frac{7 + 15}{2}\right) \)
\( = P(7,11) \)

(b)
[Diagram: Line segment PQ with points P(-3,5), R(x,y), and Q(9,-9) marked, showing midpoint calculation]
Coordinates of R are
\( R(x,y) = R\left(\frac{-3 + 9}{2}, \frac{5 - 9}{2}\right) \)
\( = R(3,-2) \)

(c)
[Diagram: Line segment MN with points M(a+b,b-a), O(x,y), and N(a-b,a+b) marked, showing midpoint calculation]
Coordinates of O are
\( O(x,y) = O\left(\frac{a+b + a-b}{2}, \frac{b-a + a+b}{2}\right) \)
\( = O(a,b) \)

(d)
[Diagram: Line segment AB with points A(3a-2b,5a+7b), C(x,y), and B(a+4b,a-3b) marked, showing midpoint calculation]
Coordinates of C are
\( C(x,y) = C\left(\frac{3a-2b + a+4b}{2}, \frac{5a+7b + a-3b}{2}\right) \)
\( = C(2a+b, 3a+2b) \)

(e)
[Diagram: Line segment PQ with points P(a+3,5b), R(x,y), and Q(3a-1,3b+4) marked, showing midpoint calculation]
Coordinates of R are
\( R(x,y) = R\left(\frac{a+3 + 3a-1}{2}, \frac{5b + 3b+4}{2}\right) \)
\( = R(2a+1, 4b+2) \)

In simple words: To find the midpoint, we add the x-coordinates and divide by 2, then add the y-coordinates and divide by 2. It is like finding the middle point between two places.

📝 Teacher's Note: Draw two points on the board and show students how to find the exact middle. Use real examples like finding the middle house between two addresses on a street.

🎯 Exam Tip: Always write the midpoint formula first. Then substitute the values. Check your arithmetic twice - small calculation mistakes lose marks.

Answer 17.
Answer:
[Diagram: Line segment AC with points A(2,5), B(1,0), and C(x,y) marked with ratios 1:2]
Given AC : AB = 3 : 1
\( \therefore \) AB : BC = 1 : 2
Coordinates of B are
\( 1 = \frac{x + 4}{3} \), \( 0 = \frac{y + 10}{3} \)
\( 3 = x + 4 \), \( 0 = y + 10 \)
\( x = -1 \), \( y = -10 \)
Hence the coordinates of C are (-1, -10).

In simple words: We use the section formula to find where point B divides the line AC. We know the ratio and coordinates of A and B, so we can find C.

📝 Teacher's Note: Show students that when we know two points and the ratio, we can find the third point. Draw this on the board with simple numbers first.

🎯 Exam Tip: Write the section formula clearly. Substitute values step by step. Always state your final answer clearly with coordinates in brackets.

Answer 18.
Answer:
[Diagram: Line segment AB with points A(2,7), Q(x,y), and B(7,12) marked with ratio 4:1]
AQ : BQ = 4 : 1
Coordinates of Q are
\( Q(x,y) = Q\left(\frac{4 \times 7 + 1 \times 2}{4+1}, \frac{4 \times 12 + 1 \times 7}{4+1}\right) = Q(6,11) \)
Thus the coordinates of Q are (6, 11).
\( AQ = \sqrt{(2-6)^2 + (7-11)^2} = \sqrt{16+16} = 4\sqrt{2} \)
\( BQ = \sqrt{(7-6)^2 + (12-11)^2} = \sqrt{1+1} = \sqrt{2} \)
\( \Rightarrow AQ = 4BQ \)

In simple words: First we find where Q divides AB using the section formula. Then we check our answer by calculating the actual distances AQ and BQ to verify the 4:1 ratio.

📝 Teacher's Note: Explain that we can verify our answer by checking if the distances really match the given ratio. This builds confidence in the solution.

🎯 Exam Tip: Always verify your answer when possible. Show the verification step - examiners give extra marks for this checking process.

Answer 19.
Answer:
[Diagram: Triangle OPQ with vertices O(0,0), P(-6,9), Q(12,-3), and internal points M and N dividing sides in ratio 1:2]
It is given that M divides OP in the ratio 1: 2 and point N divides OQ in the ratio 1: 2.
Using section formula, the coordinates of M are
\( \left(\frac{-6+0}{3}, \frac{9+0}{3}\right) = (-2,3) \)
Using section formula, the coordinates of N are
\( \left(\frac{12+0}{3}, \frac{-3+0}{3}\right) = (4,-1) \)
Thus, the coordinates of M and N are (-2, 3) and (4, -1) respectively.
Now, using distance formula, we have:
\( PQ = \sqrt{(-6-12)^2 + (9+3)^2} = \sqrt{324+144} = \sqrt{468} \)
\( MN = \sqrt{(4+2)^2 + (-1-3)^2} = \sqrt{36+36} = \sqrt{52} \)
It can be observed that:
\( PQ = \sqrt{468} = \sqrt{9 \times 52} = 3\sqrt{52} = 3MN \)
Hence, proved.

In simple words: We find points M and N using the section formula. Then we calculate distances PQ and MN. We discover that PQ is exactly 3 times MN, which proves the relationship.

📝 Teacher's Note: This shows a beautiful property - when you take points that divide two sides of a triangle in the same ratio, the line joining them is parallel to the third side and one-third its length.

🎯 Exam Tip: Write "Hence, proved" at the end of proof questions. Show all working clearly - each step should lead logically to the next.

Answer 22.
Answer:
[Diagram: Line segment AB with points A(-3,10), P(x,0), Q(0,y), and B(6,-5) on coordinate axes]
Let the coordinates of two points x-axis and y-axis be P(x, 0) and Q(0, y) respectively. Let P divides AB in the ratio k: 1.
Coordinates of P are
\( P(x,0) = P\left(\frac{6k-3}{k+1}, \frac{-5k+10}{k+1}\right) \)
\( \Rightarrow 0 = \frac{-5k+10}{k+1} \)
\( \Rightarrow 5k = 10 \)
\( \Rightarrow k = 2 \)
Hence P divides AB in the ratio 2:1.
Let Q divides AB in the ratio k₁:1.
Coordinates of Q are
\( Q(0,y) = Q\left(\frac{6k₁-3}{k+1}, \frac{-5k₁+10}{k+1}\right) \)
\( \Rightarrow 0 = \frac{6k₁-3}{k₁+1} \)
\( \Rightarrow 6k₁ = 3 \)
\( \Rightarrow k₁ = \frac{1}{2} \)
Hence Q divides AB in the ratio 1:2
Hence proved, P and Q are the points of trisection.

In simple words: We find that P divides AB in ratio 2:1 and Q divides AB in ratio 1:2. This means P and Q cut the line AB into three equal parts, so they are trisection points.

📝 Teacher's Note: Trisection means cutting into three equal parts. Show students that when one point divides in 1:2 and another in 2:1, they create three equal segments.

🎯 Exam Tip: For trisection problems, always find both ratios and verify they are 1:2 and 2:1. This confirms the points divide the line into three equal parts.

Answer 23.
Answer:
[Diagram: Line segment AB with points A(0,-4), P(x,0), Q(0,y), and B(5,8) showing trisection points on axes]
Let P(x, 0) lies on the line y = 0 i.e. x-axis and divides the line segment AB in the ratio k: 1.
Coordinates of P are
\( P(x,0) = P\left(\frac{5k-0}{k+1}, \frac{8k-4}{k+1}\right) \)
\( \Rightarrow 0 = \frac{8k-4}{k+1}, \frac{5k-0}{k+1} = x \)
\( \Rightarrow 8k = 4, \frac{5k-10}{k+1} = x \)
\( \Rightarrow k = \frac{1}{2} \ldots(1), x = -5 \)
Hence P(-5,0) divides AB in the ratio 1:2.
Let Q(0,y) lies on the line x = 0 i.e. y-axis and divides the line segment AB in the ratio k₁:1.
Coordinates of Q are
\( Q(0,y) = Q\left(\frac{5k₁-0}{k₁+1}, \frac{8k₁-4}{k₁+1}\right) \)
\( 0 = \frac{5k₁-0}{k₁+1}, y = \frac{8k₁-4}{k₁+1} \)
\( \Rightarrow 5k₁ = 0, y = \frac{8(2)-4}{2+1} \) (from(2))
\( \Rightarrow k₁ = 2 \ldots(2) y = 4 \)
Hence, Q(0,4) divides in the ratio 2:1.
Hence proved P and Q are the points of trisection of AB.

In simple words: Point P on x-axis divides AB in ratio 1:2, and point Q on y-axis divides AB in ratio 2:1. Together they cut AB into three equal parts, making them trisection points.

📝 Teacher's Note: When a line segment intersects both coordinate axes, these intersection points often trisect the segment. This is a useful property to remember.

🎯 Exam Tip: Always check that the two ratios are 1:2 and 2:1 for trisection. Write coordinates clearly and show which point lies on which axis.

Answer 2.
Answer: Coordinates of P are, \( P(6,3) = P\left(\frac{-2 + x}{2}, \frac{0 + y}{2}\right) \)
\( 6 = \frac{-2 + x}{2}, \quad 3 = \frac{y}{2} \)
\( \Rightarrow 12 = -2 + x, \quad y = 6 \)
\( \Rightarrow x = 14 \)
Coordinates of B are (14,6).
In simple words: P is the midpoint of line AB. We use the midpoint formula to find the missing point B. We put known values and solve for x and y.

[Diagram: A line segment with points A(-2,0), P(6,3), and B(x,y) marked, where P is the midpoint between A and B.]

📝 Teacher's Note: Show students how midpoint formula works. If P is exactly in the middle, then the x-coordinate of P is the average of x-coordinates of A and B. Same for y-coordinates.

🎯 Exam Tip: Always write the midpoint formula first. Then substitute the given values. Show each step clearly to get full marks.

Answer 3.
Answer: Let A(x,0) lies on x-axis and B(0,y) lies on y-axis, given AP : PB = 1 : 1
Coordinates of P are, \( P(4,-3) = P\left(\frac{x + 0}{2}, \frac{0 + y}{2}\right) \)
\( 4 = \frac{x}{2}, \quad -3 = \frac{y}{2} \)
\( x = 8, \quad y = -6 \)
Co-ordinates of A are (8,0) and B are (0,-6)
In simple words: Point A is on x-axis so its y-coordinate is 0. Point B is on y-axis so its x-coordinate is 0. P divides AB equally, so we use midpoint formula.

[Diagram: A coordinate plane showing points A(8,0) on x-axis, B(0,-6) on y-axis, and P(4,-3) as their midpoint.]

📝 Teacher's Note: Explain that points on x-axis have y = 0, and points on y-axis have x = 0. This makes the problem easier to solve.

🎯 Exam Tip: When a point lies on x-axis, write y = 0. When it lies on y-axis, write x = 0. This is the key to solving such problems.

Answer 4.
Answer: Given PQ = PR, i.e. PQ : QR = 1 : 1
Coordinates of Q are, \( Q(y,7) = Q\left(\frac{1-5}{2}, \frac{-3+x}{2}\right) \)
\( y = -2, \quad 7 = \frac{-3+x}{2} \)
\( y = -2, \quad 14 = -3+x \)
\( 17 = x \)
The values of x and y are 17 and -2 respectively.
In simple words: Q is the midpoint of line PR. We use the midpoint formula. The x-coordinate of Q is the average of x-coordinates of P and R.

[Diagram: A line segment showing points P(-5,x), Q(y,7), and R(1,-3) where Q is the midpoint.]

📝 Teacher's Note: When PQ = QR, it means Q is exactly in the middle of P and R. So Q is the midpoint of PR.

🎯 Exam Tip: Write "Q is the midpoint of PR" first. Then use the midpoint formula. Always solve for both x and y separately.

Answer 5.
Answer: \( \frac{AB}{AC} = \frac{1}{2} \)
∴ AB : BC = 1 : 1
Coordinates of B are, \( B(-2,b) = B\left(\frac{-4+a}{2}, \frac{-4+2}{2}\right) \)
\( -2 = \frac{-4+a}{2}, \quad b = -1 \)
\( -4 = -4+a, \quad b = -1 \)
The values of a and b are 0 and -1 respectively
In simple words: Since AB : AC = 1 : 2, this means AB : BC = 1 : 1. So B is the midpoint of AC. We use midpoint formula to find a and b.

[Diagram: A line segment with points A(-4,-4), B(x,y), and C(a,2) where B divides AC in ratio 1:1.]

📝 Teacher's Note: Help students understand that if AB : AC = 1 : 2, then AB : BC = 1 : 1. This means B is the midpoint of AC.

🎯 Exam Tip: First find what ratio B divides AC in. If it is 1:1, then B is the midpoint. Use the right formula for the given ratio.

Answer 6.
Answer: Given : PR : RQ = 1 : 1
Coordinates of R are, \( R(3,5) = R\left(\frac{2+n}{2}, \frac{m+4}{2}\right) \)
\( 3 = \frac{2+n}{2}, \quad 5 = \frac{m+4}{2} \)
\( 6 = 2+n, \quad 10 = m+4 \)
\( n = 4, m = 6 \)
The values of m and n are 6 and 4 respectively.
In simple words: R is the midpoint of line PQ. We use the midpoint formula to find the unknown coordinates m and n.

[Diagram: A line segment showing points P(2,m), R(3,5), and Q(n,4) where R is the midpoint.]

📝 Teacher's Note: When PR : RQ = 1 : 1, it means R divides PQ into two equal parts. So R is the midpoint of PQ.

🎯 Exam Tip: Write the midpoint formula clearly. Substitute known values and solve step by step. Show your working for both coordinates.

Answer 7.
Answer: AC : CB = 1 : 1
Coordinates of C are, \( C(2,q) = C\left(\frac{p+3}{2}, \frac{2+6}{2}\right) \)
\( 2 = \frac{p+3}{2}, \quad q = 4 \)
\( 4 = p+3, q = 4 \)
\( p = 1, q = 4 \)
The values of p and q are 1 and 4 respectively.
In simple words: C is the midpoint of line AB. We use the midpoint formula to find p and q values.

[Diagram: A line segment with points A(p,2), C(2,q), and B(3,6) where C is the midpoint.]

📝 Teacher's Note: When AC : CB = 1 : 1, point C divides AB into two equal parts. This makes C the midpoint of AB.

🎯 Exam Tip: Always check if the ratio is 1:1. If yes, use midpoint formula. If not, use section formula with the given ratio.

Answer 8.
Answer: Let O(x,y) be the centre of the circle with diameter AB, ∴ O is midpoint of AB
i.e. AO : OB = 1 : 1
Coordinates of O are, \( O(x,y) = O\left(\frac{3+7}{2}, \frac{1+11}{2}\right) = O(5,6) \)
Thus, the coordinates of centre are (5,6).
In simple words: When AB is the diameter of a circle, the center O is exactly in the middle of AB. So O is the midpoint of AB.

[Diagram: A circle with diameter AB, where A(3,1) and B(7,11) are endpoints, and O(x,y) is the center.]

📝 Teacher's Note: Teach students that the center of a circle is always the midpoint of any diameter. This is a very important property to remember.

🎯 Exam Tip: When a line is the diameter of a circle, its midpoint is the center. Use midpoint formula directly.

Answer 9.
Answer: O is the centre of the circle with diameter AB.
∴ AO : OB = 1 : 1
Coordinates of O are, \( O(3,6) = O\left(\frac{1+x}{2}, \frac{4+y}{2}\right) \)
\( 3 = \frac{1+x}{2}, \quad 6 = \frac{4+y}{2} \)
\( 6 = 1+x, \quad 12 = 4+y \)
\( x = 5, y = 8 \)
Coordinates of B are (5,8)
Length of AB = \( \sqrt{(5-1)^2 + (8-4)^2} \)
= \( \sqrt{16+16} \)
= \( 4\sqrt{2} \text{ units} \)
In simple words: The center O is the midpoint of diameter AB. We find B using midpoint formula. Then we find the length using distance formula.

[Diagram: A circle with center O(3,6), point A(1,4), and point B(x,y) where AB is the diameter.]

📝 Teacher's Note: First find the missing endpoint using midpoint formula. Then use distance formula to find the length of diameter.

🎯 Exam Tip: Write both formulas clearly - midpoint formula first, then distance formula. Show all steps including the final simplified answer.

Answer 10.
Answer: We know that in a parallelogram, diagonals bisect each other.
∴ midpoint of AC = midpoint of BD
\( O\left(\frac{6+8}{2}, \frac{-2+6}{2}\right) = O\left(\frac{x+3}{2}, \frac{y-2}{2}\right) \)
∴ \( \frac{6+8}{2} = \frac{x+3}{2}, \frac{-2+6}{2} = \frac{y-2}{2} \)
\( 14 = x+3, \quad 4 = y-2 \)
\( x = 11, y = 6 \)
The coordinates of the fourth vertex D are (11,6)
In simple words: In a parallelogram, the diagonals cut each other exactly in half. So midpoint of AC equals midpoint of BD.

[Diagram: A parallelogram with vertices A(6,-2), B(3,-2), C(8,6), and D(x,y), showing diagonals AC and BD intersecting at point O.]

📝 Teacher's Note: Explain that diagonals of a parallelogram always bisect each other. This property helps us find missing vertices easily.

🎯 Exam Tip: Write "diagonals of a parallelogram bisect each other" first. Then equate the midpoints of both diagonals.

Answer 11.
Answer: Coordinates of midpoint of PR are \( \left(\frac{-2-4}{2}, \frac{5+3}{2}\right) \) i.e. (-3,4)
Coordinates of midpoint of QS are \( \left(\frac{-9+3}{2}, \frac{2+6}{2}\right) \) i.e. (-3,4)
The midpoint of PR is same as that of QS, i.e. diagonals PR and QS bisect each other.
Hence, PQRS is a parallelogram.
In simple words: We find the midpoint of both diagonals PR and QS. Since both midpoints are the same, the diagonals bisect each other. This proves PQRS is a parallelogram.

[Diagram: A quadrilateral PQRS with vertices P(-2,5), Q(3,6), R(-4,3), and S(-9,2) showing diagonals PR and QS.]

📝 Teacher's Note: To prove a quadrilateral is a parallelogram, show that diagonals bisect each other. This means both diagonals have the same midpoint.

🎯 Exam Tip: Find midpoint of both diagonals using the midpoint formula. If they are equal, then it is a parallelogram. State the conclusion clearly.

Answer 12.
Answer: Let the coordinates of C and D be (x,y) and (a,b) respectively
Midpoint of AC is O coordinates of O are,
\( O(-3,2) = O\left(\frac{4+x}{2}, \frac{2+y}{2}\right) \)
\( -3 = \frac{4+x}{2}, 2 = \frac{2+y}{2} \)
\( -6 = 4+x, \quad 4 = 2+y \)
\( x = -10, y = 2 \)
C(-10,2)
Similarly, coordinates of midpoint of DB, i.e. O are,
\( O(-3,2) = O\left(\frac{a-1}{2}, \frac{b+5}{2}\right) \)
\( -3 = \frac{a-1}{2}, 2 = \frac{b+5}{2} \)
\( -6 = a-1, 4 = b+5 \)
\( a = -5, b = -1 \)
D(-5,-1)
Thus, the coordinates of the other two vertices are (-10,2) and (-5,-1)
In simple words: In a parallelogram, diagonals bisect each other at O. We use this fact and the midpoint formula to find the missing vertices C and D.

[Diagram: A parallelogram with known vertices A(4,2) and B(-1,5), unknown vertices C(x,y) and D(a,b), and center O(-3,2).]

📝 Teacher's Note: Since O is the midpoint of both diagonals AC and BD, we can use the midpoint formula twice to find both missing vertices.

🎯 Exam Tip: Use the property that diagonals of a parallelogram bisect each other. Apply midpoint formula for both diagonals separately.

Answer 13.
Answer: We know that in a parallelogram diagonals bisect each other
\( \therefore \) midpoint of AC = midpoint of BD

Step 1: Find midpoint of AC using coordinates A(8,5) and C(-5,5)
\( O\left(\frac{8-5}{2}, \frac{5+5}{2}\right) = O\left(\frac{x-7}{2}, \frac{y-5}{2}\right) \)

Step 2: Equate the coordinates
\( \frac{8-5}{2} = \frac{x-7}{2}, \frac{5+5}{2} = \frac{y-5}{2} \)
\( \frac{3}{2} = \frac{x-7}{2}, 10 = y-5 \)

Step 3: Solve for x and y
x = 10, y = 15

Coordinates of fourth vertex D are (10,15)

In simple words: In a parallelogram, the diagonals cross exactly at the middle. So we find where the diagonal AC crosses, and this must be the same middle point for diagonal BD.

[Diagram: This shows a parallelogram with vertices A(8,5), B(-7,-5), C(-5,5), and the unknown vertex D, with both diagonals drawn crossing at the center point.]

📝 Teacher's Note: Draw the parallelogram on graph paper first. Show students that diagonals always cut each other in half. This property makes finding the fourth vertex easy.

🎯 Exam Tip: Always write "diagonals bisect each other" first. Then use midpoint formula. Show each step clearly to get full marks.

Answer 14.
Answer: Let A(\( x_1,y_1 \)), B(\( x_2,y_2 \)) and C(\( x_3,y_3 \)) be the coordinates of the vertices ΔABC.
Midpoint of AB, i.e. D
\( D(2,1) = D\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \)

Step 1: Set up equations from midpoint coordinates
\( 2 = \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} = -1 \)
\( x_1 + x_2 = 4 - -(1), y_1 + y_2 = -2 - -(2) \)

Similarly,
\( x_1 + x_3 = -2 - -(3), y_1 + y_3 = 8 - -(4) \)
\( x_2 + x_3 = -4 - -(5), y_2 + y_3 = 4 - -(6) \)

Step 2: Solve the system of equations
Adding (1), (3) and (5): \( 2(x_1 + x_2 + x_3) = -2 \)
\( x_1 + x_2 + x_3 = -1 \)
\( 4 + x_3 = -1 \) [from(1)]
\( x_3 = -5 \)

Adding (2),(4) and (6): \( 2(y_1 + y_2 + y_3) = 10 \)
\( y_1 + y_2 + y_3 = 5 \)
\( -2 + y_3 = 5 \) [from(2)]
\( y_3 = 7 \)

Step 3: Find other coordinates
From (3): \( x_1 - 5 = -2 \), \( x_1 = 3 \)
From (4): \( y_1 + 7 = 8 \), \( y_1 = 1 \)
From (5): \( x_2 - 5 = -4 \), \( x_2 = 1 \)
From (6): \( y_2 + 7 = 4 \), \( y_2 = -3 \)

The coordinates of the vertices of ΔABC are (3,1), (1,-3) and (-5,7)

In simple words: We know the midpoints of all three sides. We use the fact that each midpoint is exactly halfway between two vertices. This gives us six equations to solve for the three vertices.

[Diagram: This shows a triangle with vertices A, B, C and the three midpoints D(2,1), E(-1,4), and F(-2,2) marked on the sides.]

📝 Teacher's Note: Show students that if you know all three midpoints, you can work backwards to find the original triangle. Each midpoint gives two equations.

🎯 Exam Tip: Write the midpoint formula first for each side. Set up all six equations clearly. Solve systematically by adding equations. Check your answer by verifying the midpoints.

Answer 15.
Answer: Let P(\( x_1, y_1 \)), Q(\( x_2, y_2 \)) and R(\( x_3, y_3 \)) be the coordinates of the vertices of ΔPQR.
Midpoint of PQ is D
\( D(-2,3) = D\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \)

\( \frac{x_1 + x_2}{2} = -2, \frac{y_1 + y_2}{2} = 3 \)
\( x_1 + x_2 = -4 - -(1), y_1 + y_2 = 6 - -(2) \)

similarly,
\( x_2 + x_3 = 8 - -(3), y_2 + y_3 = -6 - -(4) \)
\( x_1 + x_3 = 8 - -(5), y_1 + y_3 = 10 - -(6) \)

Step 1: Solve for x-coordinates
Adding (1), (3) and (5): \( 2(x_1 + x_2 + x_3) = 12 \)
\( x_1 + x_2 + x_3 = 6 \)
\( -4 + x_3 = 6 \)
\( x_3 = 10 \)

Step 2: Solve for y-coordinates
Adding (2), (4) and (6): \( 2(y_1 + y_2 + y_3) = 10 \)
\( y_1 + y_2 + y_3 = 5 \)
\( 6 + y_3 = 5 \)
\( y_3 = -1 \)

The coordinates of vertex R are (10, -1)

In simple words: We know two midpoints and one vertex. We use the midpoint formula to find where the third vertex must be. Each midpoint tells us about two vertices.

[Diagram: This shows triangle PQR with vertex P at the top, and midpoints D(-2,3) and E(4,5) marked on sides PQ and PR respectively, with F(4,-3) on side QR.]

📝 Teacher's Note: When you know one vertex and two midpoints, you can find the other two vertices step by step. Each midpoint equation gives you information about two vertices.

🎯 Exam Tip: Write down all midpoint equations first. Label them (1), (2), etc. Add the equations to eliminate variables systematically. Always check your final answer.

Answer 16.
Answer: let A(\( x_1, y_1 \)), B(\( x_2, y_2 \)) and C(\( x_3, y_3 \)) be the coordinates of the vertices of ΔABC.
D is the midpoint of AB
\( D(-3,2) = D\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \)

\( \frac{x_1 + x_2}{2} = -3, \frac{y_1 + y_2}{2} = 2 \)
\( x_1 + x_2 = -6 - - - (1) \)
\( y_1 + y_2 = 4 - - - (2) \)

Similarly
\( x_2 + x_3 = 2 - - -(3) \)
\( y_2 + y_3 = -4 - - -(4) \)
\( x_1 + x_3 = 10 - - -(5) \)
\( y_1 + y_3 = 12 - - -(6) \)

Step 1: Solve for x-coordinates
Adding (1), (3) and (5): \( 2(x_1 + x_2 + x_3) = 6 \)
\( x_1 + x_2 + x_3 = 3 \)
\( -6 + x_3 = 3 \)
\( x_3 = 9 \)

Step 2: Find other x-coordinates
From (3): \( x_2 + 9 = 2 \), \( x_2 = -7 \)
From (5): \( x_1 + 9 = 10 \), \( x_1 = 1 \)

Step 3: Solve for y-coordinates
Adding (2), (4) and (6): \( 2(y_1 + y_2 + y_3) = 12 \)
\( y_1 + y_2 + y_3 = 6 \)
\( 4 + y_3 = 6 \)
\( y_3 = 2 \)

Step 4: Find other y-coordinates
from(4): \( y_2 + 2 = -4 \), \( y_2 = -6 \)
from(6): \( y_1 + 2 = 12 \), \( y_1 = 10 \)

The coordinates of the vertices of ΔABC are (9,2), (1,10) and (-7,-6)

In simple words: We know all three midpoints of the triangle sides. We write equations for each midpoint and solve them together to find where the three corners of the triangle are.

[Diagram: This shows triangle ABC with all three midpoints marked: D(-3,2) on side AB, F(5,6) on side AC, and E(1,-2) on side BC.]

📝 Teacher's Note: This is the complete version of finding vertices from midpoints. Students often make arithmetic mistakes, so check each step carefully. Make them verify by calculating midpoints back.

🎯 Exam Tip: Set up all six equations systematically. Add equations in groups to find each vertex coordinate. Always verify your answer by checking that the midpoints are correct.

Answer 17.
Answer:
We know that the median of triangle bisects the opposite side

\( \therefore \) BD : DC = 1 : 1

Coordinates of D are,
\( D(x, y) = D\left(\frac{5+3}{2}, \frac{3-1}{2}\right) = D(4, 1) \)

Length of median AD = \( \sqrt{(7 - 4)^2 + (-3 - 1)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \) units

In simple words: A median is a line from a corner of a triangle to the middle of the opposite side. We find the middle point first, then measure the distance.

📝 Teacher's Note: Show students that the median always cuts the opposite side into two equal parts. Use real triangles cut from paper to demonstrate this.

🎯 Exam Tip: Always write "median bisects the opposite side" first. Then use the midpoint formula to find D, and distance formula for length.

Answer 18.
Answer:
Let O be the centroid of \( \triangle ABC \).
Coordinates of O are
\( O(x, y, z) = O\left(\frac{3 + 10 - 7}{3}, \frac{-5 + 4 - 2}{3}\right) \)

\( = O(2, -1) \)

In simple words: The centroid is the point where all three medians meet. It is like the balance point of the triangle. We add all x-coordinates and divide by 3, then do the same for y-coordinates.

📝 Teacher's Note: Tell students that centroid is the "center of gravity" of a triangle. If you cut a triangle from cardboard, it will balance on the centroid point.

🎯 Exam Tip: The centroid formula is always "add all coordinates and divide by 3". Write this formula first, then substitute values.

Answer 19.
Answer:
Given the centroid of \( \triangle ABC \) is at origin, i.e. G(0,0).
Let the coordinates of third vertex be (x,y).
Coordinates of G are,

\( G(0, 0) = G\left(\frac{1 + 3 + x}{3}, \frac{4 + 1 + y}{3}\right) \)

\( 0 = \frac{4 + x}{2}, 0 = \frac{5 + y}{2} \)

x = -4, y = -5
Coordinates of third vertex are (-4,-5)

In simple words: We know where the centroid is (at origin). We use this to find the missing corner of the triangle by working backwards with the centroid formula.

📝 Teacher's Note: This is like solving a puzzle. We know the answer (centroid at origin) and two pieces (given vertices), so we find the missing piece.

🎯 Exam Tip: When centroid is given, set up the centroid formula equal to the given coordinates. Then solve the two equations separately for x and y.

Answer 20.
Answer:
let G(a,b) be at centroid of \( \triangle ABC \),
Coordinates of G are,
\( G(a, b) = G\left(\frac{4 - 2 + 1}{3}, \frac{2 - 6 + 1}{3}\right) = G(1, -1) \)

Let CE be the median through C
\( \therefore \) AE : EB = 1 : 1
Coordinates of E are

\( E(x, y) = E\left(\frac{4 - 2}{2}, \frac{2 - 6}{2}\right) = E(1, -2) \)

Length of median CE
\( = \sqrt{(1 - 1)^2 + (2 - 1)^2} \)
\( = \sqrt{9} \)
\( = 3 \) units

In simple words: First we find the centroid using all three vertices. Then we find the midpoint E of side AB. Finally we measure the distance from C to E.

📝 Teacher's Note: Remind students that a median goes from any vertex to the midpoint of the opposite side. There are three medians in every triangle.

🎯 Exam Tip: Always find the midpoint first, then calculate distance. Show both steps clearly to get full marks.

Answer 21.
Answer:
let G(x,y) be the centroid of \( \triangle PQR \)
Coordinates of G are,
\( G(x, y) = G\left(\frac{5 + 3 + 1}{3}, \frac{1 + 4 + 1}{3}\right) \)

\( = G(3, 2) \)

In simple words: We find the centroid by adding all x-coordinates and dividing by 3, then doing the same for y-coordinates.

📝 Teacher's Note: The centroid divides each median in the ratio 2:1. The longer part is always towards the vertex.

🎯 Exam Tip: Use the formula \( G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \). Show this formula first, then substitute.

Answer 22.
Answer:
Let G be the centroid of \( \triangle PQR \) whose coordinates are (2,5) and let (x,y) be the coordinates of vertex P.

Coordinates of G are,
\( G(2, 5) = G\left(\frac{x - 6 + 7}{3}, \frac{y + 5 + 8}{3}\right) \)

\( 2 = \frac{x + 1}{3}, 5 = \frac{y + 13}{3} \)

6 = x + 1, 15 = y + 13

x = 5, y = 2
Coordinates of vertex P are (5,2)

In simple words: We know where the centroid is and two vertices. We work backwards to find the missing vertex using the centroid formula.

📝 Teacher's Note: This is the reverse problem of finding centroid. Students should practice both forward and backward calculations.

🎯 Exam Tip: Set up two separate equations - one for x-coordinates and one for y-coordinates. Solve each equation to find x and y.

Answer 23.
Answer:
Let G be the centroid of \( \triangle ABC \) whose coordinates are (0,-3) and let C(x,y) be the coordinates of third vertex

Coordinates of G are,
\( G(0, -3) = G\left(\frac{-1 + 5 + x}{3}, \frac{4 + 2 + y}{3}\right) \)

\( 0 = \frac{4 + x}{3}, -3 = \frac{6 + y}{3} \)

x = -4, y = -15
Coordinates of third vertex are (-4,-15)

In simple words: We know the centroid position and two corners of the triangle. We use the centroid formula to find the third missing corner.

📝 Teacher's Note: Emphasize that the centroid is always inside the triangle. If the answer seems wrong, check the calculation.

🎯 Exam Tip: Always write "Let C(x,y) be the coordinates of third vertex" first. This shows the examiner what you are finding.

Answer 24.
Answer:
Let ABC be a triangle
The midpoint of whose sides AC, AB and BC are D, E and F respectively.
We know that the centroid of \( \triangle DEF \). Let G(x,y) be the centroid of \( \triangle ABC \) and \( \triangle DEF \)
Coordinates of centroid G are,
\( G(x, y) = G\left(\frac{1 + 3 + 2}{3}, \frac{2 + 4 - 3}{3}\right) \)

\( = G(2, 1) \)

In simple words: When we connect the midpoints of a triangle's sides, we get a smaller triangle inside. The centroids of both triangles are at the same point.

📝 Teacher's Note: This is a special property - the triangle formed by joining midpoints has the same centroid as the original triangle. Draw this on the board to show students.

🎯 Exam Tip: State the property first: "The centroid of triangle ABC and triangle DEF (formed by midpoints) coincides." This shows understanding.

Answer 25.
Answer:
Let D, E and F be the midpoints of the sides AB, AC and BC of \( \triangle ABC \) respectively.

\( \therefore \) AD : DB = BF : FC = AE : EC = 1 : 1
Coordinates of D are,
\( D(x, y) = D\left(\frac{0 - 4}{2}, \frac{2 + 2}{2}\right) = D(-2, 2) \)

Similarly,
\( E(a, b) = E\left(\frac{-4 - 2}{2}, \frac{2 - 4}{2}\right) = E(-3, -1) \)

and,
\( F(p, q) = F\left(\frac{0 - 2}{2}, \frac{2 - 4}{2}\right) = F(-1, -1) \)

Coordinates of centroid of \( \triangle ABC \) are,
\( = \left(\frac{-4 - 2 + 0}{3}, \frac{2 - 4 + 2}{3}\right) = (-2, 0) \)

Coordinates of centroid of \( \triangle DEF \) are,
\( = \left(\frac{-2 - 3 - 1}{3}, \frac{2 - 1 - 1}{3}\right) = (-2, 0) \)

Thus, the centroid of \( \triangle ABC \) and \( \triangle DEF \) coincides with the centroid of \( \triangle DEF \)

In simple words: We find the midpoints of all three sides first. Then we calculate the centroid of both triangles. They turn out to be the same point.

📝 Teacher's Note: This proves the theorem that the medial triangle (formed by midpoints) has the same centroid as the original triangle. Students should verify this with different triangles.

🎯 Exam Tip: Find all three midpoints first, then calculate both centroids separately. State the conclusion clearly: "Both centroids coincide at the same point."

Answer 26.
Answer:
[Diagram: A quadrilateral ABCD with vertices A(-5,4), B(-1,-2), C(5,2), and D(x,y), with diagonals AC and BD intersecting at point O]

Given: A(-5,4), B(-1,-2), C(5,2), D(x,y)

Step 1: Find the lengths of sides AB and BC.
\[ AB = \sqrt{(-1-(-5))^2 + (-2-4)^2} = \sqrt{16 + 36} = \sqrt{52} \text{ units} \]
\[ BC = \sqrt{(-1-5)^2 + (-2-2)^2} = \sqrt{36 + 16} = \sqrt{52} \text{ units} \]

Step 2: Find the length of diagonal AC.
\[ AC = \sqrt{(5+5)^2 + (2-4)^2} = \sqrt{100 + 4} = \sqrt{104} \]

Step 3: Check if triangle ABC is right-angled.
\[ AB^2 + BC^2 = 52 + 52 = 104 \]
\[ AC^2 = 104 \]
Since \( AB^2 + BC^2 = AC^2 \), triangle ABC is a right-angled triangle with AB = BC.

Step 4: For ABCD to be a square, diagonals must bisect each other.
Let the coordinates of D be (x,y). If ABCS is a square, then:
Midpoint of AC = Midpoint of BD

\[ \left(\frac{-5+5}{2}, \frac{4+2}{2}\right) = \left(\frac{x-1}{2}, \frac{y-2}{2}\right) \]

\[ O = \left(\frac{x-1}{2}, 3\right) = \left(\frac{y-2}{2}\right) \]

Since O = (0, 3), we get:
\[ \frac{x-1}{2} = 0 \text{ and } \frac{y-2}{2} = 3 \]

Solving: x = 1, y = 8

Therefore, coordinates of D are (1,8)
In simple words: We found that AB and BC are equal and form a right angle. For a square, the diagonals must cut each other exactly in half. Using this rule, we found where D must be.

📝 Teacher's Note: Show students that in a square, all sides are equal and diagonals bisect each other at right angles. Use graph paper to plot the points and see the square shape.

🎯 Exam Tip: Always check AB = BC and use the midpoint formula for diagonals. Write "coordinates of D are (1,8)" clearly as the final answer.

Answer 27.
Answer:
[Diagram: A circle with center O(a+2,a-1), point A(2,-2) on the circle, and point B(8,-2) also on the circle]

Given: Circle with center O(a+2, a-1), points A(2,-2) and B(8,-2) on the circle

Since A and B are on the same circle with center O:
OA = OB [radii of same circle]

Step 1: Use the condition \( OA^2 = OB^2 \)
\[ (a+2-2)^2 + (a-1+2)^2 = (a+2-8)^2 + (a-1+2)^2 \]
\[ a^2 + (a+1)^2 = (a-6)^2 + (a+1)^2 \]

Step 2: Simplify the equation.
\[ a^2 = a^2 + 36 - 12a \]
\[ 12a = 36 \]
\[ a = 3 \]

Therefore, a = 3
In simple words: Since both points are on the same circle, they must be the same distance from the center. We used this fact to find the value of a.

📝 Teacher's Note: Draw a circle and mark two points on it. Show students that both points are exactly the same distance from the center. This distance is called the radius.

🎯 Exam Tip: Write "OA = OB" first, then use the distance formula. Square both sides to avoid square roots. Always write "a = 3" as your final answer.

Answer 28.
Answer:
[Diagram: Triangle ABC with vertices A(-a,0), B(0,a), and C(a,b), with centroid G marked]

Given: Triangle with vertices A(-a,0), B(0,a), C(a,b)

Step 1: Find coordinates of centroid G.
\[ G(x,y) = G\left(\frac{-a+0+a}{3}, \frac{0+a+b}{3}\right) = G\left(0, \frac{a+b}{3}\right) \]

Step 2: Calculate \( GA^2 \).
\[ GA^2 = (0-(-a))^2 + \left(\frac{a+b}{3}-0\right)^2 \]
\[ GA^2 = \frac{9a^2 + a^2 + b^2 + 2ab}{9} = \frac{10a^2 + b^2 + 2ab}{9} \]

Step 3: Calculate \( GB^2 \).
\[ GB^2 = (0-0)^2 + \left(\frac{a+b}{3}-a\right)^2 \]
\[ GB^2 = \left(\frac{b-2a}{3}\right)^2 = \frac{b^2 + 4a^2 - 4ab}{9} \]

Step 4: Calculate \( GC^2 \).
\[ GC^2 = (0-a)^2 + \left(\frac{a+b}{3}-b\right)^2 \]
\[ GC^2 = a^2 + \left(\frac{a-2b}{3}\right)^2 = \frac{9a^2 + a^2 + 4b^2 - 4ab}{9} \]

Step 5: Find \( GA^2 + GB^2 + GC^2 \).
\[ GA^2 + GB^2 + GC^2 = \frac{10a^2 + b^2 + 2ab + b^2 + 4a^2 - 4ab + 10a^2 + 4b^2 - 4ab}{9} \]
\[ = \frac{24a^2 + 6b^2 - 6ab}{9} = \frac{1}{3}(8a^2 + 2b^2 - 2ab) \]

Step 6: Calculate sides of triangle.
\[ AB^2 = (-a-0)^2 + (0-a)^2 = 2a^2 \]
\[ BC^2 = (0-a)^2 + (a-b)^2 = a^2 + a^2 + b^2 - 2ab = 2a^2 + b^2 - 2ab \]
\[ AC^2 = (-a-a)^2 + (0-b)^2 = 4a^2 + b^2 \]

\[ AB^2 + BC^2 + AC^2 = 2a^2 + 2a^2 + b^2 - 2ab + 4a^2 + b^2 = 8a^2 + 2b^2 - 2ab \]

Step 7: Prove the relationship.
From equations (1) and (2):
\[ GA^2 + GB^2 + GC^2 = \frac{1}{3}(AB^2 + BC^2 + AC^2) \]

Therefore, we have proved that \( GA^2 + GB^2 + GC^2 = \frac{1}{3}(AB^2 + BC^2 + AC^2) \)
In simple words: The centroid is the point where all three medians meet. This formula shows a special relationship between distances from centroid to vertices and the side lengths of the triangle.

📝 Teacher's Note: The centroid divides each median in ratio 2:1. Use cardboard triangles to show students where the centroid is located. It's the balance point of the triangle.

🎯 Exam Tip: Use the centroid formula correctly: add all x-coordinates and divide by 3, same for y-coordinates. Show all steps clearly and write the final relationship as given in the question.

Answer 29.
Answer:
[Diagram: Triangle with vertices A(2,5), B(-2,4), and C(-2,6)]

Given: A(2,5), B(-2,4), C(-2,6)

Step 1: Calculate the lengths of all three sides.
\[ AB = \sqrt{(2+2)^2 + (5-4)^2} = \sqrt{16 + 1} = \sqrt{17} \text{ units} \]

\[ BC = \sqrt{(-2+2)^2 + (4-6)^2} = \sqrt{0 + 4} = 2 \text{ units} \]

\[ AC = \sqrt{(2+2)^2 + (5-6)^2} = \sqrt{16 + 1} = \sqrt{17} \text{ units} \]

Step 2: Check the relationship between sides.
It can be seen that AB = AC.

Therefore, the given coordinates are the vertices of an isosceles triangle.
In simple words: An isosceles triangle has two sides of equal length. Here AB and AC are both √17 units long, so this triangle has two equal sides.

📝 Teacher's Note: Draw the triangle on graph paper. Students can see that two sides are equal length. Point out that BC is the base and is different from the other two sides.

🎯 Exam Tip: Calculate all three side lengths using distance formula. Check which sides are equal. Write "isosceles triangle" clearly as your answer since two sides are equal.