ML Aggarwal Class 11 Maths Solutions Chapter 10 Sequences and Series

Introduction

Co-ordinate geometry, also called analytical geometry, is the branch of Mathematics that uses algebra to study geometry. René Descartes, a French mathematician, realised around 1637 that a straight line or a curve in a plane can be shown by an algebraic equation. Because of this discovery, a new branch of Mathematics called Co-ordinate Geometry was born. In co-ordinate geometry, we show a point in a plane by an ordered pair of real numbers, called co-ordinates of the point; and a straight line or a curve by an algebraic equation with real coefficients. In this way, we use algebra to study straight lines and geometric curves and learn about their nature and properties.

10.1 Recapitulation

Let us go over some basic ideas of co-ordinate geometry that you studied in earlier classes.

Co-ordinate system

When two numbered lines that are perpendicular to each other (usually horizontal and vertical) are placed together so that their origins (the points where zero is marked) meet, the setup that results is called a cartesian co-ordinate system or a co-ordinate plane. If X'OX and Y'OY are two numbered lines that meet at right angles at O (shown in Fig. 10.1), then:

• (i) X'OX is called the x-axis (or axis of x).
• (ii) Y'OY is called the y-axis (or axis of y).
• (iii) X'OX and Y'OY taken together are called co-ordinate axes or rectangular axes (rectangular because they are at right angles to each other).
• (iv) The point O is called the origin.

Co-ordinates of a point

Let P be any point in the co-ordinate plane. From P, draw a perpendicular PM to X'OX (shown in Fig. 10.2), then:

• (i) OM is called the x-coordinate (or abscissa) of P and is usually shown by x.
• (ii) MP is called the y-coordinate (or ordinate) of P and is usually shown by y.
• (iii) x and y taken together are called the cartesian coordinates or simply co-ordinates of the point P and are shown by (x, y).

The position of each point on the plane is uniquely determined with reference to rectangular axes by means of an ordered pair of real numbers, called co-ordinates of the point. On the other hand, for every ordered pair of real numbers we can find a unique point on the plane. So there is a one-one match between the set of points on a plane and the set of ordered pairs of real numbers.

Distance between two points

The distance between the points P(x₁, y₁) and Q(x₂, y₂) is given by:

PQ = \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

Section formula

The co-ordinates of the point which divides the line segment joining the points P(x₁, y₁) and Q(x₂, y₂) internally in the ratio m : n are given by:

\( \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right) \)

The co-ordinates of the point which divides the line segment joining the points P(x₁, y₁) and Q(x₂, y₂) externally in the ratio m : n are given by:

\( \left( \frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n} \right) \)

In particular, the co-ordinates of the mid-point of the line segment PQ are:

\( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)

Centroid of a triangle

The co-ordinates of the centroid of a triangle whose vertices are A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are given by:

\( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \)

Incentre of a triangle

Definition: The point where any two internal angle bisectors of a triangle meet is called the incentre of the triangle. It is usually shown by I.

Note: If the internal bisector of ∠A of a triangle ABC meets the side BC at D, then \( \frac{BD}{DC} = \frac{AB}{AC} \) (from Geometry).

Find the incentre of a triangle when its vertices are known. Then prove that the internal bisectors of the angles of a triangle meet at a single point.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of triangle ABC, and let a, b, c be the lengths of the sides of triangle ABC opposite to the vertices A, B, C respectively.

Let the internal bisector of ∠A meet the side BC at D, then:

\( \frac{BD}{DC} = \frac{AB}{AC} = \frac{c}{b} \) ...(i)

This means D divides segment BC internally in the ratio c : b.

So the co-ordinates of D are:

\( \left( \frac{cx_3 + bx_2}{c + b}, \frac{cy_3 + by_2}{c + b} \right) \)

Let the internal bisector of ∠B meet AD at I, so I is the incentre of triangle ABC.

In triangle ABD, BI is the internal bisector of ∠B:

\( \frac{DI}{IA} = \frac{BD}{AB} = \frac{BD}{c} \) ...(ii)

From (i):

\( \frac{BD}{DC} = \frac{c}{b} \)
\( \implies \frac{BD}{BD + DC} = \frac{c}{c + b} \)
\( \implies \frac{BD}{BC} = \frac{c}{c + b} \)
\( \implies BD = \frac{ac}{c + b} \)

Substituting this value of BD in (ii), we get:

\( \frac{DI}{IA} = \frac{ac/(c+b)}{c} = \frac{a}{c + b} \)

This means I divides segment DA internally in the ratio a : (c + b).

Therefore, by the section formula, the co-ordinates of I are:

\( \left( \frac{ax_1 + (c + b) \cdot \frac{cx_3 + bx_2}{c + b}}{a + c + b}, \frac{ay_1 + (c + b) \cdot \frac{cy_3 + by_2}{c + b}}{a + c + b} \right) \)
\( = \left( \frac{ax_1 + bx_2 + cx_3}{a + b + c}, \frac{ay_1 + by_2 + cy_3}{a + b + c} \right) \)

The symmetry of the co-ordinates of I shows that it also lies on the internal bisector of ∠C. So the internal bisectors of the angles of a triangle meet at a single point.

Area of a triangle

Find the area of a triangle when its vertices are known.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of triangle ABC. Draw perpendiculars AM, BN and CL from these points to the x-axis, then:

NM = OM - ON = x₁ - x₂
ML = OL - OM = x₃ - x₁
NL = OL - ON = x₃ - x₂

Now, area of triangle ABC = area of trapezium ABNM + area of trapezium AMLC - area of trapezium BNLC

\( = \frac{1}{2}(NB + MA) \cdot NM + \frac{1}{2}(MA + LC) \cdot ML - \frac{1}{2}(NB + LC) \cdot NL \)

(Since area of a trapezium = \( \frac{1}{2} \) (sum of the parallel sides) × height)

\( = \frac{1}{2}[(y_2 + y_1)(x_1 - x_2) + (y_1 + y_3)(x_3 - x_1) - (y_2 + y_3)(x_3 - x_2)] \)
\( = \frac{1}{2}[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] \)

Since the area of a triangle can never be negative, we must take the absolute value.

Therefore, area of triangle ABC = \( \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \)

Corollary: Condition for collinearity of three points

Three points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) will be collinear if and only if the area of triangle ABC = 0, that is, if and only if:

\( \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0 \)

\( \implies x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0 \)

Locus of a point

The locus of a point on a plane is the curve or the path that the point traces out as it moves under a given geometric condition (or conditions).

Example 1. The base of an equilateral triangle with side 2a lies along the y-axis such that the midpoint of the base is at the origin. Find the vertices of the triangle.
Answer: Let ABC be the given equilateral triangle with side 2a whose base, say AB, lies along the y-axis and the midpoint of the base is at the origin. Two different triangles satisfy these conditions. The third vertex C can be on the right of AB or on the left of AB. Since ABC is an equilateral triangle, ∠BAC = 60°. Also, OA = \( \frac{1}{2} \times 2a = a \). From triangle AOC, \( \tan 60° = \frac{OC}{OA} \), so \( \sqrt{3} = \frac{OC}{a} \), which gives OC = a√3. Therefore, the vertices of the triangle are A(0, a), B(0, -a), C(a√3, 0) or C(-a√3, 0).
In simple words: Place two vertices on the y-axis at (0, a) and (0, -a). The third vertex can be found using the fact that in an equilateral triangle, all sides have equal length. Using the angle 60° and basic trigonometry, the third vertex lies at distance a√3 from the origin along the x-axis.

Exam Tip: When working with equilateral triangles and axes, always use the angle properties - remembering that each angle is 60°. Also keep track of the two possible positions for the third vertex.

Example 2. Find the distance between P(x₁, y₁) and Q(x₂, y₂) when (i) PQ is parallel to the y-axis (ii) PQ is parallel to the x-axis.
Answer:
(i) When PQ is parallel to the y-axis, then x₁ = x₂. So PQ = \( \sqrt{(x_1 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(y_2 - y_1)^2} = |y_2 - y_1| \).
(ii) When PQ is parallel to the x-axis, then y₁ = y₂. So PQ = \( \sqrt{(x_2 - x_1)^2 + (y_1 - y_1)^2} = \sqrt{(x_2 - x_1)^2} = |x_2 - x_1| \).
In simple words: If two points share the same x-coordinate, the distance between them is just the difference in their y-coordinates. If they share the same y-coordinate, the distance is just the difference in their x-coordinates.

Exam Tip: Remember that when one coordinate is the same for both points, the distance formula simplifies to the absolute difference of the other coordinate.

Example 3. If the distance between the points (a, -2) and (5, 1) is 5 units, find the value(s) of a.
Answer: The distance between the points (a, -2) and (5, 1) is \( \sqrt{(5 - a)^2 + (1 - (-2))^2} = \sqrt{(5 - a)^2 + 3^2} \). According to the given information, \( \sqrt{(5 - a)^2 + 9} = 5 \). Squaring both sides: (5 - a)² + 9 = 25. This gives (5 - a)² = 16, so 5 - a = ±4, which means a = 1 or a = 9. Therefore, the required values of a are 1 and 9.
In simple words: Use the distance formula and set it equal to 5. Square both sides to remove the square root, then solve for a. You will get two answers because (5 - a) can be positive or negative.

Exam Tip: Always remember to consider both positive and negative cases when taking the square root. Check both answers by substituting back into the distance formula.

Example 4. Show that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right angled triangle.
Answer: Let the points be A(7, 10), B(-2, 5) and C(3, -4). Calculate the three side lengths: AB = \( \sqrt{(-2 - 7)^2 + (5 - 10)^2} = \sqrt{81 + 25} = \sqrt{106} \); BC = \( \sqrt{(3 - (-2))^2 + (-4 - 5)^2} = \sqrt{25 + 81} = \sqrt{106} \); CA = \( \sqrt{(7 - 3)^2 + (10 - (-4))^2} = \sqrt{16 + 196} = \sqrt{212} \). So AB² = 106, BC² = 106, and CA² = 212. Since AB² + BC² = 106 + 106 = 212 = CA², the triangle is right angled with the right angle at B. Also, AB = \sqrt{106} = BC, which means the triangle is isosceles.
In simple words: Find all three side lengths using the distance formula. Check if two sides are equal (isosceles) and if the Pythagorean theorem holds (right angled).

Exam Tip: Calculate the squared distances first, not the square roots - this makes checking the Pythagorean relation much easier. Always verify both conditions: equal sides and the right angle.

Example 5. Find the point on the x-axis which is equidistant from the points (7, 6) and (3, 4).
Answer: Let P(x, 0) be any point on the x-axis. Let the given points be A(7, 6) and B(3, 4). According to the condition, AP = BP. So \( \sqrt{(x - 7)^2 + (0 - 6)^2} = \sqrt{(x - 3)^2 + (0 - 4)^2} \). Squaring: (x - 7)² + 36 = (x - 3)² + 16. Expanding: x² - 14x + 49 + 36 = x² - 6x + 9 + 16. Simplifying: -14x + 85 = -6x + 25, which gives -8x = -60, so x = 15/2. Hence, the required point is (15/2, 0).
In simple words: Set the distances from P to both given points as equal. Square to remove the square roots. Cancel out x² terms, then solve for x using simple algebra.

Exam Tip: Points equidistant from two given points always lie on the perpendicular bisector of the line joining those two points. A point on the x-axis has y-coordinate 0.

Example 6. If two vertices of an equilateral triangle are (0, 0) and (0, 2√3), find the third vertex.
Answer: Let O(0, 0), A(0, 2√3) and B(x, y) be the vertices of the equilateral triangle OAB. Calculate the distances: OA = \( \sqrt{(0 - 0)^2 + (2\sqrt{3} - 0)^2} = \sqrt{12} = 2\sqrt{3} \); OB = \( \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2} \); AB = \( \sqrt{(x - 0)^2 + (y - 2\sqrt{3})^2} = \sqrt{x^2 + y^2 - 4\sqrt{3}y + 12} \). Since the triangle is equilateral, OA = OB = AB. From OA² = OB²: 12 = x² + y² ...(i). From OA² = AB²: 12 = x² + y² - 4√3 y + 12, which gives 4√3 y = 12, so y = √3 ...(ii). Substituting into (i): x² + 3 = 12, so x² = 9, which gives x = 3 or x = -3. Hence, the third vertex is (3, √3) or (-3, √3).
In simple words: Write the equal distances condition for an equilateral triangle. Create equations by setting these distances equal. Use one equation to find y, then substitute into another equation to find x. Remember there are two possible answers.

Exam Tip: For equilateral triangles, all three sides must be equal. Setting up the three equations from pairwise distance equality usually gives enough information to solve for the unknown coordinates.

Example 7. Find the ratio in which the point P whose abscissa is 3 divides the join of A(6, 5) and B(-1, 4) and hence find the co-ordinates of P.
Answer: Let the point P divide the segment AB in the ratio k : 1. By the section formula, the co-ordinates of P are \( \left( \frac{k(-1) + 1(6)}{k + 1}, \frac{k(4) + 1(5)}{k + 1} \right) = \left( \frac{-k + 6}{k + 1}, \frac{4k + 5}{k + 1} \right) \). But the x-coordinate (abscissa) of P is 3 (given). So \( \frac{-k + 6}{k + 1} = 3 \). This gives -k + 6 = 3k + 3, so 4k = 3, which means k = 3/4. Therefore, the required ratio is 3/4 : 1, or 3 : 4 internally. The co-ordinates of P are \( \left( 3, \frac{4(3/4) + 5}{3/4 + 1} \right) = \left( 3, \frac{3 + 5}{7/4} \right) = \left( 3, \frac{32}{7} \right) \).
In simple words: Use the section formula with an unknown ratio k : 1. The x-coordinate gives you an equation to find k. Once you know k, calculate the y-coordinate using the section formula.

Exam Tip: The section formula works when you know the ratio. If you know one coordinate of the dividing point, use that to find the unknown ratio first.

Example 8. The centre of a circle is C(-2, 5) and one end of a diameter is A(3, -7), find the co-ordinates of the other end.
Answer: Let the other end of the diameter be B(α, β), where one end is A(3, -7). The midpoint of AB is \( \left( \frac{\alpha + 3}{2}, \frac{\beta - 7}{2} \right) \). Since the centre C(-2, 5) is the midpoint of AB, we have \( \frac{\alpha + 3}{2} = -2 \) and \( \frac{\beta - 7}{2} = 5 \). From the first equation: α + 3 = -4, so α = -7. From the second equation: β - 7 = 10, so β = 17. Therefore, the co-ordinates of the other end of the diameter are (-7, 17).
In simple words: The centre of a circle is the midpoint of any diameter. Use the midpoint formula. Set each coordinate of the midpoint equal to the centre's coordinates, then solve for the unknown endpoint.

Exam Tip: When a circle's centre and one endpoint of a diameter are given, use the fact that the centre bisects the diameter. The midpoint formula then gives a simple system of two linear equations.

Example 9. If three consecutive vertices of a parallelogram are (-2, -1), (1, 0) and (4, 3), find the fourth vertex.
Answer: Let A(-2, -1), B(1, 0), C(4, 3) and D(x, y) be the vertices of parallelogram ABCD. The midpoint of diagonal AC is \( \left( \frac{-2 + 4}{2}, \frac{-1 + 3}{2} \right) = (1, 1) \). The midpoint of diagonal BD is \( \left( \frac{1 + x}{2}, \frac{0 + y}{2} \right) \). Since the diagonals of a parallelogram bisect each other, the midpoints must be the same. So \( \frac{1 + x}{2} = 1 \) and \( \frac{0 + y}{2} = 1 \). This gives x = 1 and y = 2. Hence, the fourth vertex is (1, 2).
In simple words: In a parallelogram, the diagonals cut each other in half at the same midpoint. Find the midpoint of one diagonal. This midpoint must equal the midpoint of the other diagonal. Use this to find the missing vertex.

Exam Tip: The key property is that diagonals of a parallelogram bisect each other. Always calculate both midpoints and set them equal to find the fourth vertex.

Example 10. Find the co-ordinates of the incentre of the triangle whose vertices are (-2, 4), (5, 5) and (4, -2).
Answer: Let A(-2, 4), B(5, 5) and C(4, -2) be the vertices of triangle ABC. Find the side lengths: a = |BC| = \( \sqrt{(4 - 5)^2 + (-2 - 5)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \); b = |CA| = \( \sqrt{(4 - (-2))^2 + (-2 - 4)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \); c = |AB| = \( \sqrt{(5 - (-2))^2 + (5 - 4)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} \). The incentre has co-ordinates \( \left( \frac{ax_1 + bx_2 + cx_3}{a + b + c}, \frac{ay_1 + by_2 + cy_3}{a + b + c} \right) = \left( \frac{5\sqrt{2}(-2) + 6\sqrt{2}(5) + 5\sqrt{2}(4)}{5\sqrt{2} + 6\sqrt{2} + 5\sqrt{2}}, \frac{5\sqrt{2}(4) + 6\sqrt{2}(5) + 5\sqrt{2}(-2)}{16\sqrt{2}} \right) = \left( \frac{-10\sqrt{2} + 30\sqrt{2} + 20\sqrt{2}}{16\sqrt{2}}, \frac{20\sqrt{2} + 30\sqrt{2} - 10\sqrt{2}}{16\sqrt{2}} \right) = \left( \frac{40\sqrt{2}}{16\sqrt{2}}, \frac{40\sqrt{2}}{16\sqrt{2}} \right) = \left( \frac{5}{2}, \frac{5}{2} \right) \).
In simple words: Calculate the lengths of all three sides. Use the incentre formula, which weights each vertex by the length of the opposite side, divided by the sum of all side lengths.

Exam Tip: The incentre formula uses side lengths as weights. Make sure to match each side with its opposite vertex correctly - side a is opposite vertex A, and so on.

Example 11. Find the area of the triangle whose vertices are (10, -6), (2, 5) and (-1, 3).
Answer: The area of the triangle with vertices (10, -6), (2, 5) and (-1, 3) is \( \frac{1}{2}|10(5 - 3) + 2(3 - (-6)) + (-1)((-6) - 5)| = \frac{1}{2}|10(2) + 2(9) + (-1)(-11)| = \frac{1}{2}|20 + 18 + 11| = \frac{1}{2}|49| = \frac{49}{2} \) square units.
In simple words: Use the area formula. Substitute each vertex's coordinates. Simplify step by step, then take the absolute value and divide by 2.

Exam Tip: Always use absolute value in the area formula. The final area must be positive. Organize your calculation to avoid arithmetic mistakes.

Example 12. Draw the quadrilateral in the cartesian plane, whose vertices are (-4, 5), (0, 7), (5, -5) and (-4, -2). Also, find its area.
Answer: Plot the points A(-4, 5), B(0, 7), C(5, -5) and D(-4, -2) on the cartesian plane and join them to form quadrilateral ABCD. To find the area, split the quadrilateral into two triangles. Area of triangle ABC = \( \frac{1}{2}|-4(7 - (-5)) + 0((-5) - 5) + 5(5 - 7)| = \frac{1}{2}|-4(12) + 0 + 5(-2)| = \frac{1}{2}|-48 + 0 - 10| = \frac{1}{2}|-58| = 29 \) square units. Area of triangle ACD = \( \frac{1}{2}|-4((-5) - (-2)) + 5((-2) - 5) + (-4)(5 - (-5))| = \frac{1}{2}|-4(-3) + 5(-7) + (-4)(10)| = \frac{1}{2}|12 - 35 - 40| = \frac{1}{2}|-63| = \frac{63}{2} \) square units. Therefore, area of quadrilateral ABCD = 29 + 63/2 = 60 1/2 square units.
In simple words: Divide the quadrilateral into two triangles by drawing a diagonal. Calculate each triangle's area separately. Add the two areas together.

Exam Tip: Split the quadrilateral into non-overlapping triangles. Use a consistent numbering order when applying the area formula to avoid sign errors.

Example 13. For what value of x are the points (1, 5), (x, 1) and (4, 11) collinear?
Answer: The area of the triangle formed by the given points is \( \frac{1}{2}|1(1 - 11) + x(11 - 5) + 4(5 - 1)| = \frac{1}{2}|1(-10) + x(6) + 4(4)| = \frac{1}{2}|-10 + 6x + 16| = \frac{1}{2}|6x + 6| = 3|x + 1| \). The given points are collinear if and only if the area of the triangle is 0. So 3|x + 1| = 0, which gives |x + 1| = 0, meaning x + 1 = 0 (since the absolute value of a number is 0 only when the number itself is 0). Therefore, x = -1.
In simple words: Three points are collinear if the area of the triangle they form is zero. Set up the area formula, simplify, and solve for when it equals zero.

Exam Tip: Collinearity means zero area. Use the area formula and set it equal to zero. Remember that |x| = 0 only when x = 0.

Example 14. If P(x, y) is any point on the line joining the points A(a, 0) and B(0, b), then show that \( \frac{x}{a} + \frac{y}{b} = 1 \).
Answer: Since P(x, y) lies on the line joining points A(a, 0) and B(0, b), the three points P, A, B are collinear. So the area of triangle PAB is 0. Using the area formula: \( \frac{1}{2}|x(0 - b) + a(b - y) + 0(y - 0)| = 0 \). This gives |−bx + ab − ay| = 0, so −bx + ab − ay = 0. Rearranging: bx + ay = ab. Dividing by ab: \( \frac{x}{a} + \frac{y}{b} = 1 \).
In simple words: Use the fact that P is on the line AB by checking that the three points are collinear (area = 0). This collinearity condition leads directly to the required equation.

Exam Tip: This is a standard intercept form of a line. The equation shows that if a line passes through (a, 0) and (0, b), then any point on it satisfies this simple relation.

Example 15. The co-ordinates of points P, Q, R and S are (-3, 5), (4, -2), (p, 3p) and (6, 3) respectively. If the areas of triangles PQR and QRS are in the ratio 2 : 3, find p.
Answer: Area of triangle PQR = \( \frac{1}{2}|-3(-2 - 3p) + 4(3p - 5) + p(5 - (-2))| = \frac{1}{2}|6 + 9p + 12p - 20 + 7p| = \frac{1}{2}|28p - 14| = 7|2p - 1| \). Area of triangle QRS = \( \frac{1}{2}|4(3p - 3) + p(3 - (-2)) + 6((-2) - 3p)| = \frac{1}{2}|12p - 12 + 5p - 12 - 18p| = \frac{1}{2}|-p - 24| \). Given that the area ratio is 2 : 3: \( \frac{7|2p - 1|}{(1/2)|-p - 24|} = \frac{2}{3} \), which gives \( \frac{14|2p - 1|}{|-p - 24|} = \frac{2}{3} \), so \( \frac{|2p - 1|}{|-p - 24|} = \frac{1}{21} \). This means \( |2p - 1| = \frac{1}{21}|-p - 24| \). Solving: 42p - 21 = -p - 24 or 42p - 21 = p + 24, which gives 43p = -3 or 41p = 45. Therefore, p = -3/43 or p = 45/41.
In simple words: Calculate the area of each triangle. Set up the ratio equation. Solve the absolute value equation, which typically yields two cases.

Exam Tip: When working with area ratios and absolute values, remember that absolute value equations often have two solutions. Check both to see which (or both) are valid.

Example 16. If the co-ordinates of two points A, B are (1, 2), (3, 8) respectively, find a point P such that |PA| = |PB| and area of triangle PAB = 10.
Answer: Let the co-ordinates of P be (α, β). From |PA| = |PB|: PA² = PB², so (α - 1)² + (β - 2)² = (α - 3)² + (β - 8)². Expanding: α² - 2α + 1 + β² - 4β + 4 = α² - 6α + 9 + β² - 16β + 64. Simplifying: -2α - 4β + 5 = -6α - 16β + 73, which gives 4α + 12β = 68, so α + 3β = 17 ...(i). Also, area of triangle PAB = 10, so \( \frac{1}{2}|α(2 - 8) + 1(8 - β) + 3(β - 2)| = 10 \). This gives |−6α + 2 + 2β| = 20, so |−6α + 2β + 2| = 20, meaning −3α + β + 1 = 10 or −10. From −3α + β + 1 = 10: β = 3α + 9 ...(ii). From −3α + β + 1 = −10: β = 3α − 11 ...(iii). Solving (i) and (ii): α + 3(3α + 9) = 17 gives 10α = -10, so α = -1 and β = 6. Solving (i) and (iii): α + 3(3α - 11) = 17 gives 10α = 50, so α = 5 and β = 4. Hence, the point P is either (-1, 6) or (5, 4).
In simple words: First use the condition |PA| = |PB| to get one equation (the locus of points equidistant from A and B is a straight line). Then use the area condition to get another equation. Solve the system to find P.

Exam Tip: The condition |PA| = |PB| means P lies on the perpendicular bisector of AB. Combine this constraint with the area condition to locate P exactly.

Exercise 10.1

Very short answer type questions (1 to 20):

Question 1. What is the distance of the point P(x, y) from x-axis?
Answer: The distance of a point P(x, y) from the x-axis is the length of the perpendicular drawn from P to the x-axis. Since the x-axis is the line where y = 0, the foot of the perpendicular from P to the x-axis is the point (x, 0). The distance between P(x, y) and (x, 0) is |y|.
In simple words: The distance from any point to the x-axis is the absolute value of its y-coordinate.

Exam Tip: The distance to the x-axis is |y|, and the distance to the y-axis is |x|. This is a quick way to test if you understand coordinates.

Question 2. A is a point on y-axis whose ordinate is 5 and B is the point (-3, 1). Compute the length of AB.
Answer: Since A is on the y-axis with ordinate 5, its co-ordinates are (0, 5). B is the point (-3, 1). Using the distance formula: AB = \( \sqrt{(0 - (-3))^2 + (5 - 1)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \).
In simple words: Any point on the y-axis has x-coordinate 0. Substitute the co-ordinates into the distance formula.

Exam Tip: Points on the y-axis have the form (0, y). Points on the x-axis have the form (x, 0). Use these facts to quickly write coordinates.

Question 3. The distance between A(1, 3) and B(x, 7) is 5. Find the values of x.
Answer: Using the distance formula: \( \sqrt{(x - 1)^2 + (7 - 3)^2} = 5 \). Simplifying: \( \sqrt{(x - 1)^2 + 16} = 5 \). Squaring: (x - 1)² + 16 = 25, so (x - 1)² = 9. Taking square roots: x - 1 = ±3, which gives x = 4 or x = -2.
In simple words: Apply the distance formula and set it equal to 5. Square to remove the square root. Solve the resulting quadratic.

Exam Tip: When solving \(\sqrt{(x - a)^2 + b^2} = c\), always remember to consider both positive and negative square roots when you solve (x - a)² = c² - b².

Question 4. Find the abscissa of points whose ordinate is 4 and which are at a distance of 5 units from (5, 0).
Answer: Let the required point be P(x, 4). The distance from P(x, 4) to (5, 0) is 5 units. Using the distance formula: \( \sqrt{(x - 5)^2 + (4 - 0)^2} = 5 \). Simplifying: \( \sqrt{(x - 5)^2 + 16} = 5 \). Squaring: (x - 5)² + 16 = 25, so (x - 5)² = 9. Taking square roots: x - 5 = ±3, which gives x = 8 or x = 2. Therefore, the required abscissae are 8 and 2.
In simple words: The ordinate is 4, so the point is (x, 4) where x is unknown. Use the distance formula to set up an equation, then solve for x.

Exam Tip: The abscissa is the x-coordinate. If the ordinate (y-coordinate) is fixed, substitute it into the distance formula and solve for the x-coordinate.

Question 5. How many points are there on the x-axis whose distance from the point (2, 3) is less than 3 units?
Answer: Any point on the x-axis has the form (x, 0). The distance from (x, 0) to (2, 3) is \( \sqrt{(x - 2)^2 + (0 - 3)^2} = \sqrt{(x - 2)^2 + 9} \). For this distance to be less than 3 units: \( \sqrt{(x - 2)^2 + 9} < 3 \). Squaring: (x - 2)² + 9 < 9, which gives (x - 2)² < 0. But a square is never negative, so there is no value of x that satisfies this. Therefore, there are 0 points on the x-axis whose distance from (2, 3) is less than 3 units.
In simple words: Set up an inequality using the distance formula. Simplify to get (x - 2)² < 0. Since squares are always non-negative, this is impossible.

Exam Tip: The minimum distance from (2, 3) to the x-axis is 3 (the perpendicular distance). So no point on the x-axis can be closer than 3 units.

Question 6. (i) Find the point on x-axis which is equidistant from (3, 2) and (-5, -2). (ii) What point on the y-axis is equidistant from (3, 2) and (-5, -2)?
Answer: (i) Let P(x, 0) be a point on the x-axis. For P to be equidistant from A(3, 2) and B(-5, -2): PA = PB. So \( \sqrt{(x - 3)^2 + (0 - 2)^2} = \sqrt{(x - (-5))^2 + (0 - (-2))^2} \). Squaring: (x - 3)² + 4 = (x + 5)² + 4. Simplifying: (x - 3)² = (x + 5)², so x² - 6x + 9 = x² + 10x + 25, which gives -16x = 16, so x = -1. The required point is (-1, 0). (ii) Let Q(0, y) be a point on the y-axis. For Q to be equidistant from A(3, 2) and B(-5, -2): QA = QB. So \( \sqrt{(0 - 3)^2 + (y - 2)^2} = \sqrt{(0 - (-5))^2 + (y - (-2))^2} \). Squaring: 9 + (y - 2)² = 25 + (y + 2)². Expanding: 9 + y² - 4y + 4 = 25 + y² + 4y + 4. Simplifying: 13 - 4y = 29 + 4y, so -8y = 16, which gives y = -2. The required point is (0, -2).
In simple words: For (i), use a point on the x-axis. For (ii), use a point on the y-axis. In each case, set the distances to both given points as equal and solve.

Exam Tip: Points equidistant from two fixed points lie on the perpendicular bisector of the line joining them. The intersection of this perpendicular bisector with the x-axis or y-axis gives your answer.

Question 7. The points A(0, 3), B(-2, a) and C(-1, 4) are the vertices of a right angled triangle at A, find the value of a.
Answer: For the triangle to be right angled at A, the two sides AB and AC must be perpendicular. This means their dot product is zero. Vector AB = (-2 - 0, a - 3) = (-2, a - 3). Vector AC = (-1 - 0, 4 - 3) = (-1, 1). For perpendicularity: \( (-2) \times (-1) + (a - 3) \times 1 = 0 \), which gives 2 + a - 3 = 0, so a = 1.
In simple words: When an angle is a right angle, the two sides forming it are perpendicular. The dot product of the vectors along these sides must be zero.

Exam Tip: For a right angle at vertex A, use either vectors (dot product = 0) or the Pythagorean theorem (AB² + AC² = BC²).

Question 8. The centre of a circle is (2α - 1, 3α + 1) and it passes through the point (-3, -1). Find the value(s) of α if a diameter of the circle is of length 20 units.
Answer: If the circle passes through (-3, -1) and the centre is C(2α - 1, 3α + 1), then the radius is the distance from C to (-3, -1). The diameter is 20, so the radius is 10. Distance from C to (-3, -1): \( \sqrt{(2α - 1 - (-3))^2 + (3α + 1 - (-1))^2} = 10 \). Simplifying: \( \sqrt{(2α + 2)^2 + (3α + 2)^2} = 10 \). Squaring: (2α + 2)² + (3α + 2)² = 100. Expanding: 4α² + 8α + 4 + 9α² + 12α + 4 = 100. Simplifying: 13α² + 20α + 8 = 100, so 13α² + 20α - 92 = 0. Using the quadratic formula or factoring: (α + 2)(13α - 46) = 0, giving α = -2 or α = 46/13.
In simple words: The radius is half the diameter. The distance from the centre to the given point on the circle equals the radius. Set up and solve the equation.

Exam Tip: Diameter = 2 × radius. Use this to find the radius first, then apply the distance formula.

Question 9. In what ratio does the point P(1/2, 6) divide the line segment joining the points A(3, 5) and B(-7, 9)?
Answer: Let P divide AB in the ratio k : 1. By the section formula, P = \( \left( \frac{k(-7) + 1(3)}{k + 1}, \frac{k(9) + 1(5)}{k + 1} \right) = \left( \frac{-7k + 3}{k + 1}, \frac{9k + 5}{k + 1} \right) \). Since P = (1/2, 6), we have \( \frac{-7k + 3}{k + 1} = \frac{1}{2} \) and \( \frac{9k + 5}{k + 1} = 6 \). From the first equation: -14k + 6 = k + 1, so -15k = -5, giving k = 1/3. From the second equation: 9k + 5 = 6k + 6, so 3k = 1, giving k = 1/3. Both equations give k = 1/3. Therefore, P divides AB in the ratio 1/3 : 1 = 1 : 3 internally.
In simple words: Use the section formula with an unknown ratio k : 1. The x-coordinate condition and y-coordinate condition should both give the same k. Then express the ratio in simplest form.

Exam Tip: When finding the division ratio, use either the x-coordinate or y-coordinate (they must give the same answer). Then simplify the ratio into whole numbers if possible.

Question 10. Point C(-4, 1) divides the line segment joining the points A(2, -2) and B in the ratio 3 : 5. Find the point B.
Answer: Let B = (x, y). If C divides AB in the ratio 3 : 5 internally, then by the section formula: C = \( \left( \frac{3x + 5(2)}{3 + 5}, \frac{3y + 5(-2)}{3 + 5} \right) = \left( \frac{3x + 10}{8}, \frac{3y - 10}{8} \right) \). Since C = (-4, 1): \( \frac{3x + 10}{8} = -4 \) and \( \frac{3y - 10}{8} = 1 \). From the first: 3x + 10 = -32, so 3x = -42, giving x = -14. From the second: 3y - 10 = 8, so 3y = 18, giving y = 6. Therefore, B = (-14, 6).
In simple words: Given the dividing point and the ratio, use the section formula in reverse. Set up equations and solve for the unknown point.

Exam Tip: The section formula can be used in reverse. Given the dividing point, the ratio, and one endpoint, you can find the other endpoint.

Question 11. The mid-point of the line segment joining (2a, 4) and (-2, 3b) is (1, 2a + 1). Find the values of a and b.
Answer: The midpoint of the line segment joining (2a, 4) and (-2, 3b) is \( \left( \frac{2a - 2}{2}, \frac{4 + 3b}{2} \right) = \left( a - 1, \frac{4 + 3b}{2} \right) \). This equals (1, 2a + 1). So: a - 1 = 1 and \( \frac{4 + 3b}{2} = 2a + 1 \). From the first: a = 2. Substituting into the second: \( \frac{4 + 3b}{2} = 2(2) + 1 = 5 \), so 4 + 3b = 10, giving b = 2.
In simple words: Apply the midpoint formula. Equate each coordinate to the given midpoint. Solve the two equations simultaneously.

Exam Tip: The midpoint formula gives two equations (one for each coordinate). Solve them to find the unknowns.

Question 12. If the points A(-2, -1), B(1, 0), C(a, 3) and D(1, b) form a parallelogram ABCD, find the values of a and b.
Answer: In a parallelogram, the diagonals bisect each other. The diagonals are AC and BD. Midpoint of AC = \( \left( \frac{-2 + a}{2}, \frac{-1 + 3}{2} \right) = \left( \frac{a - 2}{2}, 1 \right) \). Midpoint of BD = \( \left( \frac{1 + 1}{2}, \frac{0 + b}{2} \right) = \left( 1, \frac{b}{2} \right) \). For the diagonals to bisect each other: \( \frac{a - 2}{2} = 1 \) and \( 1 = \frac{b}{2} \). From the first: a - 2 = 2, so a = 4. From the second: b = 2. Therefore, a = 4 and b = 2.
In simple words: Use the property that diagonals of a parallelogram bisect each other. Calculate both midpoints and set them equal.

Exam Tip: In a parallelogram, opposite sides are parallel and equal, and diagonals bisect each other. The midpoint property is often the quickest approach.

Question 13. The three vertices of a parallelogram taken in order are (-1, 1), (3, 1) and (2, 2) respectively. Find the coordinates of the fourth vertex.
Answer: Let A(-1, 1), B(3, 1), C(2, 2) and D(x, y) be the vertices in order. In a parallelogram, the diagonals bisect each other. The diagonals are AC and BD. Midpoint of AC = \( \left( \frac{-1 + 2}{2}, \frac{1 + 2}{2} \right) = \left( \frac{1}{2}, \frac{3}{2} \right) \). This equals the midpoint of BD = \( \left( \frac{3 + x}{2}, \frac{1 + y}{2} \right) \). So: \( \frac{3 + x}{2} = \frac{1}{2} \) and \( \frac{1 + y}{2} = \frac{3}{2} \). From the first: 3 + x = 1, so x = -2. From the second: 1 + y = 3, so y = 2. Therefore, D = (-2, 2).
In simple words: Set the midpoint of one diagonal equal to the midpoint of the other. Solve to find the missing vertex.

Exam Tip: Always use the diagonals bisect each other property for parallelograms. It's usually faster than checking parallel or equal sides.

Question 14. If the middle points of the sides of a triangle are (1, 1), (2, -3) and (3, 2), find the centroid of the triangle.
Answer: Let the vertices of the triangle be A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃). Let the midpoints be: Midpoint of BC = (1, 1), Midpoint of CA = (2, -3), Midpoint of AB = (3, 2). From these: \( \frac{x_2 + x_3}{2} = 1 \Rightarrow x_2 + x_3 = 2 \) ... (1), \( \frac{x_3 + x_1}{2} = 2 \Rightarrow x_3 + x_1 = 4 \) ... (2), \( \frac{x_1 + x_2}{2} = 3 \Rightarrow x_1 + x_2 = 6 \) ... (3). Similarly: \( \frac{y_2 + y_3}{2} = 1 \Rightarrow y_2 + y_3 = 2 \) ... (4), \( \frac{y_3 + y_1}{2} = -3 \Rightarrow y_3 + y_1 = -6 \) ... (5), \( \frac{y_1 + y_2}{2} = 2 \Rightarrow y_1 + y_2 = 4 \) ... (6). Adding (1), (2), (3): 2(x₁ + x₂ + x₃) = 12, so x₁ + x₂ + x₃ = 6. Adding (4), (5), (6): 2(y₁ + y₂ + y₃) = 0, so y₁ + y₂ + y₃ = 0. The centroid is \( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) = \left( \frac{6}{3}, \frac{0}{3} \right) = (2, 0) \).
In simple words: Write three equations from the three midpoints. Add them to find the sum of coordinates. Divide by 3 to get the centroid.

Exam Tip: When given the three midpoints, adding all three equations at once is faster than solving for individual vertices.

Question 15. If (4, -3) and (-9, 7) are two vertices of a triangle whose centroid is (1, 4), then find the third vertex.
Answer: Let the third vertex be C(x, y). The centroid of a triangle with vertices A(4, -3), B(-9, 7) and C(x, y) is given by \( \left( \frac{4 - 9 + x}{3}, \frac{-3 + 7 + y}{3} \right) = \left( \frac{x - 5}{3}, \frac{y + 4}{3} \right) \). Since the centroid is (1, 4): \( \frac{x - 5}{3} = 1 \) and \( \frac{y + 4}{3} = 4 \). From the first: x - 5 = 3, so x = 8. From the second: y + 4 = 12, so y = 8. Therefore, the third vertex is C(8, 8).
In simple words: Use the centroid formula with one unknown vertex. Set each coordinate equal to the given centroid and solve.

Exam Tip: The centroid formula averages the three vertices' coordinates. Given two vertices and the centroid, you can find the third by solving simple linear equations.

Question 16. The vertices of a triangle are A(-5, 3), B(p, -1) and C(6, q). Find the values of p and q if the centroid of the triangle ABC is the point (1, -1).
Answer: The centroid is \( \left( \frac{-5 + p + 6}{3}, \frac{3 - 1 + q}{3} \right) = \left( \frac{p + 1}{3}, \frac{q + 2}{3} \right) \). This equals (1, -1). So: \( \frac{p + 1}{3} = 1 \) and \( \frac{q + 2}{3} = -1 \). From the first: p + 1 = 3, so p = 2. From the second: q + 2 = -3, so q = -5. Therefore, p = 2 and q = -5.
In simple words: Apply the centroid formula. Equate each coordinate to the given centroid value. Solve for the unknowns.

Exam Tip: The centroid formula is straightforward: average the x-coordinates and average the y-coordinates. Always apply it systematically.

Question 17. If the point (-3, a) is the image of the point (1, a + 4) in the point (b, 1), find the values of a and b.
Answer: When a point P is the image of a point Q in a point R, the point R is the midpoint of PQ. So (b, 1) is the midpoint of (-3, a) and (1, a + 4). Using the midpoint formula: \( \frac{-3 + 1}{2} = b \) and \( \frac{a + (a + 4)}{2} = 1 \). From the first: \( \frac{-2}{2} = b \), so b = -1. From the second: \( \frac{2a + 4}{2} = 1 \), so 2a + 4 = 2, giving a = -1. Therefore, a = -1 and b = -1.
In simple words: The image of a point in another point means the second point is the midpoint. Apply the midpoint formula.

Exam Tip: "Image of P in R" means R is the midpoint of P and its image. This is a standard reflection concept.

Question 18. Show that the points (3, -2), (5, 2) and (8, 8) are collinear.
Answer: Three points are collinear if the area of the triangle they form is zero. Using the area formula for the triangle with vertices (3, -2), (5, 2) and (8, 8): Area = \( \frac{1}{2}|3(2 - 8) + 5(8 - (-2)) + 8((-2) - 2)| = \frac{1}{2}|3(-6) + 5(10) + 8(-4)| = \frac{1}{2}|-18 + 50 - 32| = \frac{1}{2}|0| = 0 \). Since the area is zero, the three points are collinear.
In simple words: Three points are collinear if they form a triangle of zero area. Use the area formula and verify it equals zero.

Exam Tip: Collinearity can be checked using the area formula or by verifying that the slope between the first two points equals the slope between the second two points.

Question 19. For what value of k are the points (8, 1), (k, -4) and (2, -5) collinear?
Answer: Three points are collinear if the area of the triangle they form is zero. Area = \( \frac{1}{2}|8(-4 - (-5)) + k((-5) - 1) + 2(1 - (-4))| = \frac{1}{2}|8(1) + k(-6) + 2(5)| = \frac{1}{2}|8 - 6k + 10| = \frac{1}{2}|18 - 6k| \). For collinearity, this must equal zero: |18 - 6k| = 0, so 18 - 6k = 0, giving k = 3. Therefore, k = 3.
In simple words: Set up the area formula with k as an unknown. Set the area equal to zero and solve for k.

Exam Tip: When an unknown appears in the area formula, set the area equal to zero and solve for the unknown to find the collinearity condition.

Question 20. Find the equation of the locus of a point which is equidistant from the points (1, 3) and (-2, 1).
Answer: Let P(x, y) be any point on the locus. The condition is that P is equidistant from A(1, 3) and B(-2, 1). So PA = PB. Squaring: PA² = PB². This gives (x - 1)² + (y - 3)² = (x + 2)² + (y - 1)². Expanding the left side: x² - 2x + 1 + y² - 6y + 9 = x² + 4x + 4 + y² - 2y + 1. Simplifying: -2x - 6y + 10 = 4x - 2y + 5. Rearranging: -6x - 4y + 5 = 0, or 6x + 4y - 5 = 0. Therefore, the equation of the locus is 6x + 4y - 5 = 0.
In simple words: Let P(x, y) be a general point. Write the equidistant condition as PA = PB. Square and expand. Simplify to get the equation.

Exam Tip: Points equidistant from two fixed points form a straight line - the perpendicular bisector of the line joining them. The algebraic method shown here derives its equation.

10.2 Shifting of Origin

Sometimes a problem involving a given set of axes can be solved more easily by moving the origin to a new point while keeping the axes parallel to the original axes. This move of the origin to a new point without changing the axes' direction is known as translation of axes. We will see how the co-ordinates of a point and the equation of a curve change under translation of axes, and how we can return to the original axes.

10.2.1 To shift the origin to a new point, without changing the direction of axes

Let OX and OY be the original axes and O' be the new origin. Let the co-ordinates of O' referred to the original axes OX and OY be (h, k). Let O'X' and O'Y' be drawn parallel to and in the same direction as OX and OY respectively. Let P be any point whose co-ordinates are (x, y) with respect to the original axes OX and OY, and (X, Y) with respect to the new axes O'X' and O'Y'. Draw PM'M parallel to OY, meeting O'X' at M' and OX at M. Extend Y'O' to meet OX at N. Then: x = OM = ON + NM = ON + O'M' = h + X = X + h, and y = MP = MM' + M'P = NO' + M'P = k + Y = Y + k. Therefore: x = X + h, y = Y + k, which gives us X = x - h, Y = y - k.

Consequences of this transformation:

• (i) A point whose co-ordinates were (x, y) now has co-ordinates (x - h, y - k).
• (ii) The co-ordinates of the old origin referred to the new axes are (-h, -k).
• (iii) A curve whose equation was f(x, y) = 0 is now shown by f(X + h, Y + k) = 0.

An obvious fact: Shifting axes cannot change lengths like the distance between two points, the distance from a point to a line, or the area of a figure. So when solving problems of this kind, you do not need to shift back to the original axes.

Example 1. Transform 2x - 3y + 5 = 0 to the parallel axes through the point (2, -3).
Answer: Let the co-ordinates (x, y) of a point on the given line change to (X, Y) when we shift the origin to (2, -3), with the new axes remaining parallel to the original axes. Then x = X + 2 and y = Y - 3. Substituting these into the given equation 2x - 3y + 5 = 0: 2(X + 2) - 3(Y - 3) + 5 = 0, which simplifies to 2X + 4 - 3Y + 9 + 5 = 0, giving 2X - 3Y + 18 = 0.
In simple words: Replace x with (X + h) and y with (Y + k) where (h, k) is the new origin. Simplify to get the new equation.

Exam Tip: When transforming equations by shifting origin, always substitute x = X + h and y = Y + k, then simplify carefully.

Example 2. Transform x² + y² - 6x + 10y - 2 = 0 to the parallel axes through (3, -5).
Answer: Let the co-ordinates (x, y) of a point on the given curve change to (X, Y) when we shift the origin to (3, -5), with the new axes remaining parallel to the original axes. Then x = X + 3 and y = Y - 5. The given equation is x² + y² - 6x + 10y - 2 = 0. Substituting: (X + 3)² + (Y - 5)² - 6(X + 3) + 10(Y - 5) - 2 = 0. Expanding: X² + 6X + 9 + Y² - 10Y + 25 - 6X - 18 + 10Y - 50 - 2 = 0. Simplifying: X² + Y² + (6X - 6X) + (-10Y + 10Y) + (9 + 25 - 18 - 50 - 2) = 0, which gives X² + Y² - 36 = 0 or X² + Y² = 36.
In simple words: Substitute the transformation equations into the original equation and simplify. The new equation should be simpler.

Exam Tip: This transformation reveals that the original equation represents a circle with centre (3, -5) and radius 6. Shifting the origin to the centre gives the simplest form.

Example 3. Find the point to which the origin should be shifted so that the equation y² - 6y - 4x + 13 = 0 will not contain a term in y and will have no constant term.
Answer: Let the origin shift to (h, k), keeping axes parallel to the original axes. If co-ordinates (x, y) of a point change to (X, Y), then x = X + h and y = Y + k. Substituting into the given equation: (Y + k)² - 6(Y + k) - 4(X + h) + 13 = 0. Expanding: Y² + 2kY + k² - 6Y - 6k - 4X - 4h + 13 = 0. Rearranging: Y² + (2k - 6)Y - 4X + (k² - 6k - 4h + 13) = 0. For this equation to have no Y term and no constant term: 2k - 6 = 0 and k² - 6k - 4h + 13 = 0. From the first: k = 3. Substituting into the second: 9 - 18 - 4h + 13 = 0, so 4 - 4h = 0, giving h = 1. Therefore, the origin should shift to (1, 3).
In simple words: After shifting, set the coefficients of the linear terms and the constant term equal to zero. Solve for h and k.

Exam Tip: When asked to eliminate certain terms, identify which coefficients should become zero and set up equations from that condition.

Example 4. Reduce the equation x² + 4y² - 6x - 8y - 12 = 0 to the form AX² + BY² = K by shifting the origin to a suitable point.
Answer: Let the origin shift to (h, k), keeping axes parallel. If x = X + h and y = Y + k, then substituting into the given equation: (X + h)² + 4(Y + k)² - 6(X + h) - 8(Y + k) - 12 = 0. Expanding: X² + 2hX + h² + 4Y² + 8kY + 4k² - 6X - 6h - 8Y - 8k - 12 = 0. Rearranging: X² + 4Y² + (2h - 6)X + (8k - 8)Y + (h² + 4k² - 6h - 8k - 12) = 0. For the form AX² + BY² = K, we need the coefficients of X and Y to be zero: 2h - 6 = 0 and 8k - 8 = 0. This gives h = 3 and k = 1. Then: h² + 4k² - 6h - 8k - 12 = 9 + 4(1) - 6(3) - 8(1) - 12 = 9 + 4 - 18 - 8 - 12 = -25. So the equation becomes X² + 4Y² - 25 = 0 or X² + 4Y² = 25.
In simple words: Identify which coefficients must become zero in the new equation. Solve for the new origin. Then calculate the resulting constant term.

Exam Tip: The original equation can be rewritten by completing the square: (x - 3)² + 4(y - 1)² = 25. This shows directly that the origin should shift to (3, 1).

10.3 Slope of a Straight Line

Inclination of a straight line

The angle that a straight line, say L, makes with the positive direction of the x-axis, measured anticlockwise to the part of the line above the x-axis (shown in Fig. 10.9), is called the inclination (or angle of inclination) of the line L. The inclination is usually shown by θ. Clearly, 0° ≤ θ - 180°.

Particular cases:

• (i) The inclination of a line parallel to the x-axis or the x-axis itself is 0°.
• (ii) The inclination of a line parallel to the y-axis or the y-axis itself is 90°.

Horizontal, vertical and oblique lines

• (i) Any line parallel to the x-axis or the x-axis itself is called a horizontal line.
• (ii) Any line parallel to the y-axis or the y-axis itself is called a vertical line.
• (iii) A line which is neither horizontal nor vertical is called an oblique line.

Slope (or gradient) of a straight line

If θ (≠ 90°) is the inclination of a straight line, then tan θ is called its slope (or gradient). The slope of a line is usually shown by m. So if θ (≠ 90°) is the inclination of a line, then m = tan θ.

Remarks:

• 1. Since tan θ is not defined when θ = 90°, the slope of a vertical line is not defined.
• 2. Slope of the y-axis is not defined.
• 3. Since the inclination of every line parallel to the x-axis is 0°, its slope = tan 0° = 0. So the slope of every horizontal line is zero.
• 4. Slope of the x-axis is zero.

10.3.1 Slope of a straight line joining two points

To find the slope of a non-vertical line passing through two given points: Let P(x₁, y₁) and Q(x₂, y₂) be two given points on a non-vertical line l. Since line l is non-vertical, x₂ ≠ x₁. Let θ be the inclination of the line l. The inclination θ can be acute or obtuse. We consider two cases.

From P and Q, draw perpendiculars PM and QN to the x-axis and PL - NQ as shown in the figures.

Case I. When angle θ is acute:

From the figure, PL = MN = ON - OM = x₂ - x₁ and LQ = NQ - NL = NQ - MP = y₂ - y₁. In the right triangle QPL, ∠QPL = θ. So tan θ = \( \frac{LQ}{PL} = \frac{y_2 - y_1}{x_2 - x_1} \) ...(i)

Case II. When angle θ is obtuse:

From the figure, LP = NM = OM - ON = x₁ - x₂ and LQ = NQ - NL = NQ - MP = y₂ - y₁. In the right triangle QPL, ∠LPQ = π - θ. So tan(π - θ) = \( \frac{LQ}{LP} = \frac{y_2 - y_1}{x_1 - x_2} \). This gives -tan θ = \( \frac{y_2 - y_1}{x_1 - x_2} \), which means tan θ = \( \frac{y_2 - y_1}{x_2 - x_1} \).

In both cases, slope of the line l = m = tan θ = \( \frac{y_2 - y_1}{x_2 - x_1} \).

Hence, the slope m of a non-vertical line passing through the points P(x₁, y₁) and Q(x₂, y₂) is given by:

m = \( \frac{y_2 - y_1}{x_2 - x_1} \)

Example 1. Find the slope of the line passing through the points: (i) (3, -2) and (-1, 4) (ii) (3, -2) and (7, -2) (iii) (3, -2) and (3, 4)
Answer:
(i) Slope = \( \frac{4 - (-2)}{-1 - 3} = \frac{6}{-4} = -\frac{3}{2} \).
(ii) Slope = \( \frac{-2 - (-2)}{7 - 3} = \frac{0}{4} = 0 \). We can also note that the given points (3, -2) and (7, -2) have the same y-coordinate, so the line is horizontal with slope 0.
(iii) The given points (3, -2) and (3, 4) have the same x-coordinate, so the line is vertical and its slope is not defined.
In simple words: Use the slope formula. When the y-coordinates are the same, the slope is 0. When the x-coordinates are the same, the slope is undefined.

Exam Tip: Always check if the x-coordinates are the same first - if they are, the slope is undefined. Otherwise, apply the formula directly.

Exercise 10.1

Question 1. Find |y|.
Answer: |y|
In simple words: The absolute value of y is simply |y|.

Exam Tip: Absolute value gives the distance from zero, always non-negative.

Question 2. Find the distance.
Answer: 5 units
In simple words: The space between two points measures 5 units.

Exam Tip: Always verify distance calculations using the distance formula \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \).

Question 3. Find the value.
Answer: 4 or -2
In simple words: There are two possible answers: either 4 or negative 2.

Exam Tip: When you get two solutions, always check both by substituting back into the original equation.

Question 4. Find the value.
Answer: 2 or 8
In simple words: The answer can be either 2 or 8.

Exam Tip: Present all valid solutions clearly; do not omit either answer even if one seems less likely.

Question 5. Find the answer.
Answer: None
In simple words: There is no valid answer to this problem.

Exam Tip: "None" means the problem has no solution - state this explicitly if that is what your working proves.

Question 6. Find the coordinates.
(i) (−1, 0)
(ii) (0, −2)
Answer: (i) The point lies at (−1, 0).
(ii) The point lies at (0, −2).
In simple words: First point is at x = −1 and y = 0. Second point is at x = 0 and y = −2.

Exam Tip: Always write coordinates as ordered pairs (x, y) - order matters.

Question 7. Find the value.
Answer: 1
In simple words: The answer is 1.

Exam Tip: Double-check your calculations - a value of 1 is often found in coordinate geometry and slope problems.

Question 8. Find the values.
Answer: 2, −\(\frac{46}{13}\)
In simple words: One answer is 2, and the other answer is negative 46 divided by 13.

Exam Tip: When fractions appear in answers, always simplify to lowest terms and verify both solutions satisfy the original conditions.

Question 9. Find the ratio.
Answer: 1 : 3 internally
In simple words: A point divides a line segment in the ratio 1 to 3, measured from inside the segment.

Exam Tip: "Internally" means the point lies between the two endpoints; distinguish this from external division where the point lies outside.

Question 10. Find the coordinates.
Answer: (−14, 6)
In simple words: The point you are looking for is located at x = −14 and y = 6.

Exam Tip: Check your answer by verifying it satisfies any given conditions (collinearity, distance, or section formula).

Question 11. Find the values of a and b.
Answer: a = 2, b = 2
In simple words: Both a and b have the value 2.

Exam Tip: When solving simultaneous equations for two unknowns, substitute back to verify both values work in both original equations.

Question 12. Find the values of a and b.
Answer: a = 4, b = 2
In simple words: The value of a is 4, and the value of b is 2.

Exam Tip: Compare your answer with Question 11 to see how different conditions lead to different values.

Question 13. Find the coordinates.
Answer: (−2, 2)
In simple words: The point is at x = −2 and y = 2.

Exam Tip: Note that this point lies in the second quadrant (negative x, positive y).

Question 14. Find the coordinates.
Answer: (2, 0)
In simple words: The point is at x = 2 and y = 0, which means it sits on the horizontal axis.

Exam Tip: Points with y-coordinate 0 always lie on the x-axis.

Question 15. Find the coordinates.
Answer: (8, 8)
In simple words: Both x and y equal 8, so the point is at (8, 8).

Exam Tip: When x = y, the point lies on the line y = x.

Question 16. Find the values of p and q.
Answer: p = 2, q = −5
In simple words: The value p equals 2, while q equals negative 5.

Exam Tip: Always match the signs correctly when writing your final answer.

Question 17. Find the values of a and b.
Answer: a = −1, b = −1
In simple words: Both a and b equal negative 1.

Exam Tip: Negative coefficients are common in coordinate geometry - handle them carefully in your calculations.

Question 19. Find the value.
Answer: 3
In simple words: The answer is 3.

Exam Tip: Verify this result using the relevant formula or property stated in the problem.

Question 20. Find the equation of the line.
Answer: 6x + 4y - 5 = 0
In simple words: The line's equation is 6x plus 4y minus 5, all equal to zero.

Exam Tip: Always check that your line equation is in standard form with integer coefficients, and verify known points satisfy it.

Question 22. Find the coordinates.
Answer: \(\left(2 + \frac{\sqrt{11}}{2}, \frac{5}{2}\right)\) or \(\left(2 - \frac{\sqrt{11}}{2}, \frac{5}{2}\right)\)
In simple words: There are two points - both have y-coordinate 5/2, but their x-coordinates involve square root of 11.

Exam Tip: When surds appear in answers, leave them in exact form - do not convert to decimals unless specifically asked.

Question 23. Find the coordinates.
Answer: (5, 6) or (3, 4)
In simple words: There are two possible points: one at (5, 6) and another at (3, 4).

Exam Tip: When multiple solutions exist, list all of them clearly - omitting any solution leads to lost marks.

Question 24. Find the distance.
Answer: \(\sqrt{10}\) units
In simple words: The distance is the square root of 10, measured in units.

Exam Tip: Leave distance answers in surd form unless the problem asks for a decimal approximation.

Question 25. Find the ratio and coordinates.
Answer: The point divides the line segment in the ratio 3 : 2 internally, and its coordinates are \(\left(0, \frac{11}{5}\right)\).
In simple words: The dividing point is located 3 units away from one end and 2 units from the other (measured internally), and sits at x = 0, y = 11/5.

Exam Tip: Always apply the section formula correctly - the ratio tells you how to weight the endpoints in your calculation.

Question 26. Find the ratio and the value.
Answer: The point divides the line segment in the ratio 3 : 2 internally, and the answer is −\(\frac{2}{5}\).
In simple words: The line is cut in the ratio 3 to 2 from inside, and the calculated value comes to negative 2/5.

Exam Tip: When a ratio is given with a value, both parts must be calculated and shown in your working.

Question 27. Find the coordinates.
(i) \(\left(\frac{3}{2}, \frac{7}{2}\right)\)
(ii) (2, 5)
Answer: (i) The coordinates are \(\left(\frac{3}{2}, \frac{7}{2}\right)\).
(ii) The coordinates are (2, 5).
In simple words: First point is at 3/2 for x and 7/2 for y. Second point is at x = 2 and y = 5.

Exam Tip: Keep fractions in their simplest form and verify both sub-answers before finalizing.

Question 28. Find the distance.
Answer: 5 units
In simple words: The total space between the two points measures 5 units.

Exam Tip: A distance of 5 units often signals a 3-4-5 right triangle - check if your points form one.

Question 29. Find the coordinates.
Answer: (1, −12) and (5, −10)
In simple words: The two points are located at (1, −12) and (5, −10).

Exam Tip: When you find multiple points, always check that each one satisfies all the original conditions.

Question 30. Find the coordinates.
(i) (2, −5) and (12, 5)
(ii) (−5, 4) or (1, 2)
Answer: (i) The two points are (2, −5) and (12, 5).
(ii) The point is either (−5, 4) or (1, 2).
In simple words: In part (i), you get two definite points. In part (ii), there are two possible answers.

Exam Tip: Read each sub-question carefully - some ask for one point, others for multiple points or choices.

Question 31. Find the coordinates.
Answer: (−1, 0)
In simple words: The point is at x = −1 and y = 0.

Exam Tip: This point lies on the x-axis since y = 0.

Question 32. Find the values.
Answer: −3, \(\frac{21}{13}\)
In simple words: One answer is −3, and the other is the fraction 21/13.

Exam Tip: When answers involve both integers and fractions, ensure they are both in simplest form.

Question 33. Find the area.
Answer: \(\frac{1}{2}|2p^2 - 3p - 2|\) square units
In simple words: The area is one-half of the absolute value of (2p squared minus 3p minus 2), measured in square units.

Exam Tip: The absolute value ensures the area is always positive - area can never be negative.

Question 34. Find the area.
Answer: 15 square units
In simple words: The region enclosed has an area of 15 square units.

Exam Tip: Always verify area calculations using the coordinate geometry formula for triangles or quadrilaterals.

Question 36. Find the coordinates.
Answer: (7, 2) or (1, 0)
In simple words: There are two possible points: one is at (7, 2) and the other is at (1, 0).

Exam Tip: Always list both solutions when a problem yields multiple valid answers.

Exercise 10.2

Question 1. Find the coordinates.
Answer: (3, −5)
In simple words: The point is located at x = 3 and y = −5.

Exam Tip: After transformation or translation, always verify the new point satisfies the given transformation rule.

Question 2. Find the equation.
Answer: X + Y = 0
In simple words: The equation is X plus Y equals zero, or equivalently, Y = −X.

Exam Tip: The equation X + Y = 0 represents a straight line passing through the origin with slope −1.

Question 3. Find the equation.
Answer: X\(^2\) + Y\(^2\) − 13 = 0
In simple words: The equation is X squared plus Y squared minus 13, all equal to zero.

Exam Tip: This is a circle centered at the origin with radius \(\sqrt{13}\).

Question 4. Find the equation after transformation.
Answer: (X − c)\(^2\) + Y\(^2\) = r\(^2\)
In simple words: The transformed equation is (X minus c) squared plus Y squared equals r squared, which is a circle centered at (c, 0).

Exam Tip: Recognize the standard form of a circle - this shows a horizontal shift by c units and a fixed radius r.

Question 5. Find the equation.
Answer: X\(^2\) + Y\(^2\) = 50
In simple words: The equation is X squared plus Y squared equals 50, a circle centered at the origin.

Exam Tip: The radius of this circle is \(\sqrt{50} = 5\sqrt{2}\) units.

Question 6. Find the coordinates.
Answer: \(\left(\frac{3}{4}, -2\right)\)
In simple words: The point is at x = 3/4 and y = −2.

Exam Tip: Keep fractions in simplest form in your final answer.

Question 7. Find the coordinates.
Answer: (1, 1)
In simple words: The point has both x and y equal to 1.

Exam Tip: When x = y, the point always lies on the line y = x.

Exercise 10.3

Question 1. Find the slope.
(i) −\(\frac{12}{7}\)
(ii) −1
(iii) 0
(iv) not defined
Answer: (i) The slope is −\(\frac{12}{7}\).
(ii) The slope is −1.
(iii) The slope is 0 (horizontal line).
(iv) The slope is not defined (vertical line).
In simple words: Slopes can be positive, negative, zero (flat), or undefined (straight up). Calculate them using the rise-over-run formula.

Exam Tip: A slope of 0 means a horizontal line; undefined slope means a vertical line.

Question 2. Find the slope.
(i) 0
(ii) not defined
Answer: (i) The slope is 0.
(ii) The slope is not defined.
In simple words: The first line is horizontal (zero slope), and the second line is vertical (undefined slope).

Exam Tip: Always identify whether a line is horizontal or vertical before calculating its slope.

Question 3. Find the slope.
(i) \(\frac{1}{\sqrt{3}}\)
(ii) −\(\sqrt{3}\)
(iii) −1
(iv) not defined
Answer: (i) The slope is \(\frac{1}{\sqrt{3}}\).
(ii) The slope is −\(\sqrt{3}\).
(iii) The slope is −1.
(iv) The slope is not defined.
In simple words: Different angles produce different slopes - positive slopes tilt up, negative slopes tilt down, zero is flat, and undefined is vertical.

Exam Tip: Rationalize denominators when surds appear - write \(\frac{1}{\sqrt{3}}\) as \(\frac{\sqrt{3}}{3}\) in your final answer.

Question 4. Find the slope.
Answer: −\(\frac{1}{\sqrt{3}}\)
In simple words: The slope of the line is negative one divided by the square root of 3.

Exam Tip: When slopes are reciprocals and opposite in sign, the lines are perpendicular.

Question 5. Find the angle.
(i) 30°
(ii) 45°
(iii) 135°
(iv) 60°
(v) 120°
(vi) 0°
Answer: (i) The angle is 30°.
(ii) The angle is 45°.
(iii) The angle is 135°.
(iv) The angle is 60°.
(v) The angle is 120°.
(vi) The angle is 0°.
In simple words: These angles show how steep or flat a line is. Smaller angles (like 0° or 30°) are flatter; larger angles (like 120° or 135°) are steeper.

Exam Tip: Recall that slopes are related to angles by the formula: slope = tan(angle).

Question 6. Find the angle.
Answer: 135°
In simple words: The line makes an angle of 135 degrees with the positive x-axis.

Exam Tip: An angle of 135° is in the second quadrant, indicating a negative slope.

Question 7. Are the lines parallel?
Answer: No
In simple words: The two lines do not run alongside each other - they are not parallel.

Exam Tip: Parallel lines have equal slopes. If slopes differ, the lines will meet or be perpendicular.

Question 8. Find the value.
Answer: 1
In simple words: The answer is 1.

Exam Tip: Always show your working step-by-step to demonstrate how you arrived at the value.

Question 9. Find the slope.
Answer: −\(\frac{7}{3}\)
In simple words: The slope is negative 7 divided by 3.

Exam Tip: Negative slopes indicate lines that descend as you move from left to right.

Question 10. Find the slope.
Answer: −\(\frac{3}{5}\)
In simple words: The slope is negative 3 divided by 5.

Exam Tip: Keep slopes in simplest fractional form unless the problem specifies otherwise.

Exercise 10.4

Question 2. Determine the relationship between the lines.
(i) Perpendicular
(ii) Parallel
(iii) Neither
(iv) Perpendicular
Answer: (i) The lines are perpendicular to each other.
(ii) The lines are parallel to each other.
(iii) The lines are neither parallel nor perpendicular.
(iv) The lines are perpendicular to each other.
In simple words: Check whether two lines meet at a right angle (perpendicular), run side by side (parallel), or neither.

Exam Tip: Perpendicular lines have slopes whose product is −1; parallel lines have equal slopes.

Question 3. Find the value.
Answer: \(\frac{10}{3}\)
In simple words: The answer is the fraction 10 divided by 3.

Exam Tip: Always simplify fractions to their lowest form in your final answer.

Question 4. Find the value.
Answer: 9
In simple words: The answer is 9.

Exam Tip: Verify your calculation by substituting back or using an alternative method.

Question 5. Find the value.
Answer: −\(\frac{4}{3}\)
In simple words: The value is negative 4 divided by 3.

Exam Tip: Negative results are valid - do not assume all answers must be positive.

Question 6. Select the correct options.
Answer: (ii), (iv)
In simple words: The correct choices are options (ii) and (iv).

Exam Tip: For "select all that apply" questions, verify each option independently before finalizing your selections.

Question 7. Find the value.
Answer: 2
In simple words: The answer is 2.

Exam Tip: Double-check that your value makes sense in the context of the original problem.

Question 8. Find the acute angle between two lines.
Answer: The acute angle between the lines is given by tan θ = \(\frac{11}{23}\).
In simple words: When two lines cross, they make an acute angle. The tangent of this angle equals 11 divided by 23.

Exam Tip: Use the formula \(\text{tan θ} = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|\) for the angle between two lines with slopes \(m_1\) and \(m_2\).

Question 9. Find the acute angle between lines AB and BC.
Answer: The acute angle between the lines AB and BC is given by tan θ = \(\frac{2}{3}\).
In simple words: The angle where line AB meets line BC has a tangent value of 2/3.

Exam Tip: Calculate the slopes of both lines first, then apply the angle formula.

Question 11. Find the slope.
(i) −\(\frac{1}{8}\)
(ii) 8
(iii) −\(\frac{1}{8}\)
Answer: (i) The slope is −\(\frac{1}{8}\).
(ii) The slope is 8.
(iii) The slope is −\(\frac{1}{8}\).
In simple words: Part (i) and part (iii) both give the same slope of −1/8; part (ii) gives a steep positive slope of 8.

Exam Tip: When two different sub-parts yield the same answer, verify your work - it could be correct, or it could signal an error.

Question 15. Find the value and coordinates.
(i) \(\frac{7}{3}\)
(ii) (5, 8)
Answer: (i) The value is \(\frac{7}{3}\).
(ii) The coordinates are (5, 8).
In simple words: Part (i) asks for a numerical value which is 7/3. Part (ii) asks for a point location which is at x = 5, y = 8.

Exam Tip: Read each sub-part carefully - they may ask for different types of answers (numbers vs. coordinates).

Question 17. Find the relationship.
Answer: D = 2(T + 1)
In simple words: The variable D equals 2 times the quantity (T plus 1), or equivalently, D = 2T + 2.

Exam Tip: Always verify a relationship by substituting known values or checking limiting cases.