CBSE Class 10 Mathematics Coordinate Geometry Assignment Set 02

CO-ORDINATE GEOMETRY

IMPORTANT CONCEPTS

1. Distance Formula:
The distance between two points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) is given by the formula:
\(AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
COROLLARY: The distance of the point \(P(x, y)\) from the origin \(O(0, 0)\) is given by:
\(OP = \sqrt{(x - 0)^2 + (y - 0)^2}\), i.e., \(OP = \sqrt{x^2 + y^2}\)

2. Section Formula:
The coordinates of the point \(P(x, y)\) which divides the line segment joining \(A(x_1, y_1)\) and \(B(x_2, y_2)\) internally in the ratio \(m:n\) are given by:
\(x = \frac{mx_2 + nx_1}{m + n}\), \(y = \frac{my_2 + ny_1}{m + n}\)

3. Midpoint Formula:
If \(R\) is the mid-point, then \(m_1 = m_2\) and the coordinates of \(R\) are:
\(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)

4. Coordinates of the centroid of a triangle:
The coordinates of the centroid of a triangle whose vertices are \(P(x_1, y_1)\), \(Q(x_2, y_2)\), and \(R(x_3, y_3)\) are:
\(\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)\)

5. Area of a Triangle:
The area of the triangle formed by the points \(P(x_1, y_1)\), \(Q(x_2, y_2)\), and \(R(x_3, y_3)\) is the numerical value of the expression:
\(\text{ar}(\triangle PQR) = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)

LEVEL-I

Question 1. Find the distance of the points (6,-6) from origin.
Answer: Let the point be \(P(6, -6)\) and the origin be \(O(0, 0)\).
Using the distance formula from the origin, \(OP = \sqrt{x^2 + y^2}\)
\(OP = \sqrt{6^2 + (-6)^2}\)
\(OP = \sqrt{36 + 36}\)
\(OP = \sqrt{72} = 6\sqrt{2} \text{ units}\)
In simple words: The distance from the origin is found by squaring the coordinates, adding them, and taking the square root.

Exam Tip: Remember that \(\sqrt{a^2 + a^2} = a\sqrt{2}\), which saves time for points of the form \((a, a)\).

Question 2. Show that the point (1,1)(-2,7) and (3,-3) are collinear.
Answer: Let \(A(1, 1)\), \(B(-2, 7)\), and \(C(3, -3)\) be the three points.
We can show they are collinear by proving that the area of the triangle formed by them is zero.
\(\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)
Substitute the coordinates: \(x_1=1, y_1=1, x_2=-2, y_2=7, x_3=3, y_3=-3\)
\(\text{Area} = \frac{1}{2} |1(7 - (-3)) + (-2)(-3 - 1) + 3(1 - 7)|\)
\(= \frac{1}{2} |1(10) + (-2)(-4) + 3(-6)|\)
\(= \frac{1}{2} |10 + 8 - 18|\)
\(= \frac{1}{2} |18 - 18| = 0\)
Since the area of the triangle is 0, the given points are collinear.
In simple words: Three points lie on the same straight line if the triangle formed by them has zero area.

Exam Tip: Alternatively, you can show collinearity by proving that the sum of any two distances equals the third distance, or by showing that slope of AB equals slope of BC.

Question 3. Find the distance between the points R(a+b , a-b)and S(a-b, -1-b)
Answer: Let the points be \(R(a+b, a-b)\) and \(S(a-b, -1-b)\).
Using the distance formula \(RS = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\):
\(x_2 - x_1 = (a - b) - (a + b) = a - b - a - b = -2b\)
\(y_2 - y_1 = (-1 - b) - (a - b) = -1 - b - a + b = -1 - a\)
\(RS = \sqrt{(-2b)^2 + (-1 - a)^2}\)
\(RS = \sqrt{4b^2 + (a + 1)^2}\)
Note: Depending on standard textbook variations where the second point is \(S(a-b, a-b)\), let us also check if the printed answer \(2\sqrt{a^2 + b^2}\) implies a slight typo in the question coordinate. Assuming standard simplification: if \(S(a-b, -b)\), distance is \(2\sqrt{a^2+b^2}\). Following exact verbatim terms given:
\(RS = \sqrt{4b^2 + (a+1)^2}\)
In simple words: Substitute the algebraic coordinates directly into the standard distance formula and simplify.

Exam Tip: Be careful with signs when subtracting binomial expressions like \((a+b)\).

Question 4. Find the point on x-axis which is equidistant from (2,-5) and (-2,9).
Answer: Any point on the x-axis is of the form \(P(x, 0)\).
Let this point be equidistant from \(A(2, -5)\) and \(B(-2, 9)\).
Therefore, \(PA^2 = PB^2\)
\((x - 2)^2 + (0 - (-5))^2 = (x - (-2))^2 + (0 - 9)^2\)
\((x - 2)^2 + 25 = (x + 2)^2 + 81\)
\(x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81\)
Cancel \(x^2\) and \(4\) from both sides:
\(-4x + 25 = 4x + 81\)
\(25 - 81 = 4x + 4x\)
\(-56 = 8x\)
\(x = -7\)
Thus, the required point is \((-7, 0)\).
In simple words: Any point on the x-axis has a y-coordinate of zero. Equate the squared distances to solve for x easily without square roots.

Exam Tip: Always equate squares of distances (\(PA^2 = PB^2\)) instead of raw distances to avoid complicated square root radicals.

Question 5. Find the area of the triangle whose vertices(-5,-1),(3,-5)(5,2)
Answer: Let the vertices be \(A(-5, -1)\), \(B(3, -5)\), and \(C(5, 2)\).
Using the area of a triangle formula:
\(\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)
Substitute \(x_1=-5, y_1=-1, x_2=3, y_2=-5, x_3=5, y_3=2\):
\(\text{Area} = \frac{1}{2} |-5(-5 - 2) + 3(2 - (-1)) + 5(-1 - (-5))|\)
\(= \frac{1}{2} |-5(-7) + 3(3) + 5(4)|\)
\(= \frac{1}{2} |35 + 9 + 20|\)
\(= \frac{1}{2} |64| = 32 \text{ sq. units}\)
In simple words: Plug the coordinates into the triangle area formula and compute the absolute value.

Exam Tip: Area can never be negative. Always enclose the final numerical value inside modulus bars.

LEVEL-II

Question 1. Show that the points (-2,5), (3,-4) and (7,10) are the vertices of a right angled isosceles triangle.
Answer: Let \(A(-2, 5)\), \(B(3, -4)\), and \(C(7, 10)\) be the given points.
Find the squares of the lengths of the sides using the distance formula:
\(AB^2 = (3 - (-2))^2 + (-4 - 5)^2 = 5^2 + (-9)^2 = 25 + 81 = 106\)
\(BC^2 = (7 - 3)^2 + (10 - (-4))^2 = 4^2 + 14^2 = 16 + 196 = 212\)
\(CA^2 = (-2 - 7)^2 + (5 - 10)^2 = (-9)^2 + (-5)^2 = 81 + 25 = 106\)
Since \(AB^2 = CA^2\) (\(106 = 106\)), the triangle is isosceles (\(AB = CA\)).
Also, check Pythagoras theorem: \(AB^2 + CA^2 = 106 + 106 = 212 = BC^2\).
Since \(AB^2 + CA^2 = BC^2\), it is a right-angled triangle at vertex \(A\).
Thus, the given points form a right-angled isosceles triangle.
In simple words: Find the side lengths. If two sides are equal and satisfy Pythagoras theorem, it is a right-angled isosceles triangle.

Exam Tip: Working with squared lengths (\(AB^2, BC^2, CA^2\)) avoids dealing with square roots and makes applying Pythagoras theorem straightforward.

Question 2. Find a relation between x and y if the points (x,y),(1,2) and (7,0) are collinear.
Answer: Let \(A(x, y)\), \(B(1, 2)\), and \(C(7, 0)\) be collinear.
The area of the triangle formed by collinear points is zero:
\(\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0\)
\(x(2 - 0) + 1(0 - y) + 7(y - 2) = 0\)
\(2x - y + 7y - 14 = 0\)
\(2x + 6y - 14 = 0\)
Divide the entire equation by 2:
\(x + 3y = 7\)
In simple words: Set the area formula for three points equal to zero to find the linear relationship between x and y.

Exam Tip: You can also solve this by equating the slope of AB to the slope of BC.

Question 3. Find the point on y axis which is equidistance from the points (5,-2) and (-3,2)
Answer: Any point on the y-axis is of the form \(P(0, y)\).
Let this point be equidistant from \(A(5, -2)\) and \(B(-3, 2)\).
Therefore, \(PA^2 = PB^2\)
\((0 - 5)^2 + (y - (-2))^2 = (0 - (-3))^2 + (y - 2)^2\)
\(25 + (y + 2)^2 = 9 + (y - 2)^2\)
\(25 + y^2 + 4y + 4 = 9 + y^2 - 4y + 4\)
Cancel \(y^2\) and \(4\) from both sides:
\(25 + 4y = 9 - 4y\)
\(4y + 4y = 9 - 25\)
\(8y = -16\)
\(y = -2\)
Thus, the required point is \((0, -2)\).
In simple words: Points on the y-axis have an x-coordinate of zero. Use the distance squared equality to find y.

Exam Tip: Remember: x-axis points are \((x, 0)\) and y-axis points are \((0, y)\).

Question 4. If the points A(4,3) and B(x,5) are on the circle with the centre O(2,3) find the value of x.
Answer: Since \(A\) and \(B\) lie on the circle with center \(O\), the distances \(OA\) and \(OB\) are radii of the circle and are equal: \(OA = OB\)
Therefore, \(OA^2 = OB^2\)
\((4 - 2)^2 + (3 - 3)^2 = (x - 2)^2 + (5 - 3)^2\)
\(2^2 + 0^2 = (x - 2)^2 + 2^2\)
\(4 = (x - 2)^2 + 4\)
Subtract 4 from both sides:
\((x - 2)^2 = 0\)
\(x - 2 = 0 \implies x = 2\)
In simple words: The distance from the center to any point on a circle is the radius, so the distances from the center to A and B are equal.

Question 5. Find the value of 'k' for which the points (7,-2),(5,1) and (3,k) are collinear.
Answer: Since the points \((7, -2)\), \((5, 1)\), and \((3, k)\) are collinear, the area of the triangle formed by them is zero:
\(\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0\)
\(7(1 - k) + 5(k - (-2)) + 3(-2 - 1) = 0\)
\(7 - 7k + 5(k + 2) + 3(-3) = 0\)
\(7 - 7k + 5k + 10 - 9 = 0\)
\(-2k + 8 = 0\)
\(2k = 8 \implies k = 4\)
In simple words: Set the area expression to zero for collinear points and solve the simple linear equation for k.

Exam Tip: Alternatively, equate slope between first two points to slope between second and third points.

Question 6. Find the area of triangle whose vertices are (2,-4),(-1,0) and (2,4)
Answer: Let the vertices be \(A(2, -4)\), \(B(-1, 0)\), and \(C(2, 4)\).
Using the area formula:
\(\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)
\(\text{Area} = \frac{1}{2} |2(0 - 4) + (-1)(4 - (-4)) + 2(-4 - 0)|\)
\(= \frac{1}{2} |2(-4) - 1(8) + 2(-4)|\)
\(= \frac{1}{2} |-8 - 8 - 8|\)
\(= \frac{1}{2} |-24| = 12 \text{ sq. units}\)
In simple words: Substitute the coordinates into the standard triangle area formula and simplify.

Exam Tip: Double-check sign distributions inside brackets to avoid calculation errors.

Question 7. Find the ratio in which line segment joining the points (6,4) and (1,-7) is divided by x-axix also find the coordinates of the points of division.
Answer: Let the ratio be \(k:1\). The points are \(A(6, 4)\) and \(B(1, -7)\).
Using the section formula, the y-coordinate of the dividing point on the x-axis is 0:
\(y = \frac{my_2 + ny_1}{m + n}\)
\(0 = \frac{k(-7) + 1(4)}{k + 1}\)
\(0 = -7k + 4\)
\(7k = 4 \implies k = \frac{4}{7}\)
Thus, the ratio is \(4:7\) (or \(7:4\) depending on endpoint order; ratio is \(7:4\) from A to B).
Now, find the x-coordinate using \(k = \frac{4}{7}\):
\(x = \frac{mx_2 + nx_1}{m + n} = \frac{\frac{4}{7}(1) + 1(6)}{\frac{4}{7} + 1} = \frac{\frac{4}{7} + 6}{\frac{11}{7}} = \frac{\frac{4 + 42}{7}}{\frac{11}{7}} = \frac{46}{11}\)
Thus, the coordinates of the point of division are \(\left(\frac{46}{11}, 0\right)\).
In simple words: Since the x-axis divides the line, the y-coordinate is zero. Use this to find the ratio, then find the x-coordinate.

Exam Tip: Whenever a line is divided by the x-axis, set \(y = 0\); if divided by the y-axis, set \(x = 0\).

LEVEL-III

Question 1. Show that the points (7,10),(-2,5) and (3,-4) are the vertices of an isosceles right triangle.
Answer: Let \(A(7, 10)\), \(B(-2, 5)\), and \(C(3, -4)\).
Calculate the squares of the side lengths:
\(AB^2 = (-2 - 7)^2 + (5 - 10)^2 = (-9)^2 + (-5)^2 = 81 + 25 = 106\)
\(BC^2 = (3 - (-2))^2 + (-4 - 5)^2 = 5^2 + (-9)^2 = 25 + 81 = 106\)
\(CA^2 = (7 - 3)^2 + (10 - (-4))^2 = 4^2 + 14^2 = 16 + 196 = 212\)
Since \(AB^2 = BC^2\) (\(106 = 106\)), the triangle is isosceles (\(AB = BC\)).
Check Pythagoras theorem: \(AB^2 + BC^2 = 106 + 106 = 212 = CA^2\).
Since the sum of the squares of two sides equals the square of the third side, it is a right-angled triangle at vertex \(B\).
Therefore, the points form an isosceles right-angled triangle.
In simple words: Calculate side squares. Two equal sides combined with Pythagoras theorem proves it is an isosceles right triangle.

Exam Tip: Clearly state which vertex holds the 90-degree angle based on the largest side squared.

Question 2. In what ratio does the line x-y-2=0 divide the line segment joining (3,-1) and (8,9)? Also find the coordinates of the point of intersection.
Answer: Let the line \(x - y - 2 = 0\) divide the line segment joining \(A(3, -1)\) and \(B(8, 9\) in the ratio \(k:1\).
Using the section formula, the coordinates of the point of intersection \(P\) are:
\(P = \left( \frac{8k + 3}{k + 1}, \frac{9k - 1}{k + 1} \right)\)
Since point \(P\) lies on the line \(x - y - 2 = 0\), substitute these coordinates into the line equation:
\(\left(\frac{8k + 3}{k + 1}\right) - \left(\frac{9k - 1}{k + 1}\right) - 2 = 0\)
Multiply through by \((k + 1)\):
\((8k + 3) - (9k - 1) - 2(k + 1) = 0\)
\(8k + 3 - 9k + 1 - 2k - 2 = 0\)
\(-3k + 2 = 0 \implies 3k = 2 \implies k = \frac{2}{3}\)
Thus, the ratio is \(2:3\) internally.
Now, substitute \(k = \frac{2}{3}\) into the coordinates of \(P\):
\(x = \frac{8\left(\frac{2}{3}\right) + 3}{\frac{2}{3} + 1} = \frac{\frac{16}{3} + 3}{\frac{5}{3}} = \frac{\frac{25}{3}}{\frac{5}{3}} = 5\)
\(y = \frac{9\left(\frac{2}{3}\right) - 1}{\frac{2}{3} + 1} = \frac{6 - 1}{\frac{5}{3}} = \frac{5}{\frac{5}{3}} = 3\)
Thus, the coordinates of the point of intersection are \((5, 3)\).
In simple words: Express the division point in terms of ratio k, plug it into the line equation to find k, then evaluate the point coordinates.

Exam Tip: This is a standard high-weightage board question. Master substituting parametric coordinates into linear equations.

Question 3. Three consecutive vertices of a parallelogram are (-2,-1),(1,0) and (4,3). Find the coordinates of the fourth vertex.
Answer: Let the consecutive vertices be \(A(-2, -1)\), \(B(1, 0)\), \(C(4, 3)\), and let the fourth vertex be \(D(x, y)\).
In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of diagonal \(AC\) is equal to the midpoint of diagonal \(BD\).
Midpoint of \(AC\):
\(\left( \frac{-2 + 4}{2}, \frac{-1 + 3}{2} \right) = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1)\)
Midpoint of \(BD\):
\(\left( \frac{1 + x}{2}, \frac{0 + y}{2} \right)\)
Equating the midpoints:
\(\frac{1 + x}{2} = 1 \implies 1 + x = 2 \implies x = 1\)
\(\frac{0 + y}{2} = 1 \implies y = 2\)
Thus, the coordinates of the fourth vertex are \((1, 2)\).
In simple words: Use the property that diagonals of a parallelogram bisect each other to equate their midpoints.

Exam Tip: Ensure vertices are taken in consecutive order along the perimeter when naming A, B, C, D.

Question 4. Show that the points A(5,6); B(1,5); C(2,1) and D(6,2) are the vertices of a square.
Answer: Let the given points be \(A(5, 6)\), \(B(1, 5)\), \(C(2, 1)\), and \(D(6, 2)\).
Find the squares of all four side lengths and two diagonals:
\(AB^2 = (1 - 5)^2 + (5 - 6)^2 = (-4)^2 + (-1)^2 = 16 + 1 = 17\)
\(BC^2 = (2 - 1)^2 + (1 - 5)^2 = 1^2 + (-4)^2 = 1 + 16 = 17\)
\(CD^2 = (6 - 2)^2 + (2 - 1)^2 = 4^2 + 1^2 = 16 + 1 = 17\)
\(DA^2 = (5 - 6)^2 + (6 - 2)^2 = (-1)^2 + 4^2 = 1 + 16 = 17\)
Since \(AB = BC = CD = DA = \sqrt{17}\), all four sides are equal.
Now, check the diagonals:
\(AC^2 = (2 - 5)^2 + (1 - 6)^2 = (-3)^2 + (-5)^2 = 9 + 25 = 34\)
\(BD^2 = (6 - 1)^2 + (2 - 5)^2 = 5^2 + (-3)^2 = 25 + 9 = 34\)
Since the four sides are equal and the diagonals are equal (\(AC = BD\)), the given points form a square.
In simple words: A quadrilateral with all four sides equal and equal diagonals is a square.

Exam Tip: Proving all four sides equal is not enough for a square; you must also prove both diagonals are equal.

Question 5. The vertices of a triangle are (-1,3),(1,-1) and (5,1). Find the length of medians through vertices (-1,3) and (5,1)
Answer: Let the vertices be \(A(-1, 3)\), \(B(1, -1)\), and \(C(5, 1)\).
1. Median through vertex \(A(-1, 3)\):
This median meets the opposite side \(BC\) at its midpoint \(D\).
Coordinates of \(D\) (midpoint of \(BC\)):
\(D = \left( \frac{1 + 5}{2}, \frac{-1 + 1}{2} \right) = \left( \frac{6}{2}, \frac{0}{2} \right) = (3, 0)\)
Length of median \(AD\):
\(AD = \sqrt{(3 - (-1))^2 + (0 - 3)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\)
2. Median through vertex \(C(5, 1)\):
This median meets the opposite side \(AB\) at its midpoint \(E\).
Coordinates of \(E\) (midpoint of \(AB\)):
\(E = \left( \frac{-1 + 1}{2}, \frac{3 + (-1)}{2} \right) = \left( \frac{0}{2}, \frac{2}{2} \right) = (0, 1)\)
Length of median \(CE\):
\(CE = \sqrt{(0 - 5)^2 + (1 - 1)^2} = \sqrt{(-5)^2 + 0^2} = \sqrt{25} = 5\)
Thus, the lengths of both medians are 5 and 5 (or formatted as \((5, 5)\)).
In simple words: Find the midpoint of the opposite side, then calculate the distance from the vertex to that midpoint.

Exam Tip: A median connects a vertex to the midpoint of the opposite side.

Question 6. Find the value of P for which the points (-5,1),(1,P) and (4,-2) are collinear.
Answer: Since the points \((-5, 1)\), \((1, P)\), and \((4, -2)\) are collinear, the area of the triangle formed by them is zero:
\(\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0\)
\(-5(P - (-2)) + 1(-2 - 1) + 4(1 - P) = 0\)
\(-5(P + 2) + 1(-3) + 4(1 - P) = 0\)
\(-5P - 10 - 3 + 4 - 4P = 0\)
\(-9P - 9 = 0\)
\(-9P = 9 \implies P = -1\)
In simple words: Set the area formula to zero for collinear points and solve the linear equation for P.

Exam Tip: Collinear points always yield zero triangular area.

SELF EVALUATION QUESTION

Question 1. Find the distance between points.
a. A(6,0) B(14,0)
b. A(0,-5) B(0,10)
c. A(0,p) B(P,0)
Answer:
(a) Distance between \(A(6, 0)\) and \(B(14, 0)\):
\(AB = \sqrt{(14 - 6)^2 + (0 - 0)^2} = \sqrt{8^2} = 8 \text{ units}\)
(b) Distance between \(A(0, -5)\) and \(B(0, 10)\):
\(AB = \sqrt{(0 - 0)^2 + (10 - (-5))^2} = \sqrt{15^2} = 15 \text{ units}\)
(c) Distance between \(A(0, p)\) and \(B(p, 0)\):
\(AB = \sqrt{(p - 0)^2 + (0 - p)^2} = \sqrt{p^2 + p^2} = \sqrt{2p^2} = |p|\sqrt{2} \text{ units}\)
In simple words: When points lie on the same coordinate axis, the distance is simply the absolute difference of their non-zero coordinates.

Exam Tip: For points on axes, you can skip the full distance formula and directly subtract coordinates.

Question 2. Show that the points (-1,-1),(1,1) and (-\(\sqrt{3}\), \(\sqrt{3}\)) are the vertices of an equilateral triangle.
Answer: Let \(A(-1, -1)\), \(B(1, 1)\), and \(C(-\sqrt{3}, \sqrt{3})\).
Calculate the squares of all three side lengths:
\(AB^2 = (1 - (-1))^2 + (1 - (-1))^2 = 2^2 + 2^2 = 4 + 4 = 8\)
\(BC^2 = (-\sqrt{3} - 1)^2 + (\sqrt{3} - 1)^2 = (3 + 2\sqrt{3} + 1) + (3 - 2\sqrt{3} + 1) = 8\)
\(CA^2 = (-1 - (-\sqrt{3}))^2 + (-1 - \sqrt{3})^2 = (\sqrt{3} - 1)^2 + (-1 - \sqrt{3})^2 = (3 - 2\sqrt{3} + 1) + (3 + 2\sqrt{3} + 1) = 8\)
Since \(AB^2 = BC^2 = CA^2 = 8\), all three sides are equal (\(AB = BC = CA = \sqrt{8} = 2\sqrt{2}\)).
Therefore, the points form an equilateral triangle.
In simple words: A triangle with all three sides equal in length is an equilateral triangle.

Exam Tip: Proving all side lengths equal is both necessary and sufficient for an equilateral triangle.

Question 3. The line joining the points A(4,-5) and B(4,5) is divided by the point P such that AP/AB=2/5. Find the coordinates of P.
Answer: Given \(\frac{AP}{AB} = \frac{2}{5}\). This means if \(AB = 5\) units, then \(AP = 2\) units.
Consequently, \(PB = AB - AP = 5 - 2 = 3\) units.
Thus, point \(P\) divides the line segment \(AB\) internally in the ratio \(AP:PB = 2:3\).
Using the section formula with \(A(4, -5)\) and \(B(4, 5)\) and ratio \(m:n = 2:3\):
\(x = \frac{2(4) + 3(4)}{2 + 3} = \frac{8 + 12}{5} = \frac{20}{5} = 4\)
\(y = \frac{2(5) + 3(-5)}{2 + 3} = \frac{10 - 15}{5} = \frac{-5}{5} = -1\)
Thus, the coordinates of \(P\) are \((4, -1)\).
In simple words: Convert the ratio AP to AB into the ratio AP to PB, then apply the section formula.

Exam Tip: Be careful: if given \(AP:AB\), always subtract to find \(PB\) before applying the section formula.

Question 4. Find the coordinates of the points which trisect the line segment joining (1,-2) and (-3,4).
Answer: Let the endpoints be \(A(1, -2)\) and \(B(-3, 4)\).
Let the two points of trisection be \(P\) and \(Q\).
1. Point \(P\) divides \(AB\) in the ratio \(1:2\):
\(P = \left( \frac{1(-3) + 2(1)}{1 + 2}, \frac{1(4) + 2(-2)}{1 + 2} \right) = \left( \frac{-3 + 2}{3}, \frac{4 - 4}{3} \right) = \left( -\frac{1}{3}, 0 \right)\)
2. Point \(Q\) divides \(AB\) in the ratio \(2:1\):
\(Q = \left( \frac{2(-3) + 1(1)}{2 + 1}, \frac{2(4) + 1(-2)}{2 + 1} \right) = \left( \frac{-6 + 1}{3}, \frac{8 - 2}{3} \right) = \left( -\frac{5}{3}, 2 \right)\)
Thus, the coordinates of the points of trisection are \(\left(-\frac{1}{3}, 0\right)\) and \(\left(-\frac{5}{3}, 2\right)\).
In simple words: Trisection points divide the line segment in ratios \(1:2\) and \(2:1\).

Exam Tip: Trisection means dividing into three equal parts, which requires two division points.

Question 5. Determine the ratio in which the line 2x+y=4 divides the line segment joining the points (2,-2) and (3,7).
Answer: Let the line \(2x + y = 4\) divide the line segment joining \(A(2, -2)\) and \(B(3, 7)\) in the ratio \(k:1\).
Using the section formula, the coordinates of the point of division \(P\) are:
\(P = \left( \frac{3k + 2}{k + 1}, \frac{7k - 2}{k + 1} \right)\)
Since point \(P\) lies on the line \(2x + y = 4\), substitute these coordinates into the line equation:
\(2\left(\frac{3k + 2}{k + 1}\right) + \left(\frac{7k - 2}{k + 1}\right) = 4\)
Multiply through by \((k + 1)\):
\(2(3k + 2) + (7k - 2) = 4(k + 1)\)
\(6k + 4 + 7k - 2 = 4k + 4\)
\(13k + 2 = 4k + 4\)
\(13k - 4k = 4 - 2\)
\(9k = 2 \implies k = \frac{2}{9}\)
Thus, the required ratio is \(2:9\) internally.
In simple words: Set up the general division point using ratio k, substitute it into the line equation, and solve for k.

Exam Tip: If k turns out positive, the division is internal; if negative, it is external.

Question 6. Find the value of K such that the point (0,2) is equidistant from the points (3,k) and (k,5).
Answer: Let \(P(0, 2)\) be equidistant from \(A(3, k)\) and \(B(k, 5)\).
Therefore, \(PA^2 = PB^2\)
\((3 - 0)^2 + (k - 2)^2 = (k - 0)^2 + (5 - 2)^2\)
\(9 + (k^2 - 4k + 4) = k^2 + 3^2\)
\(9 + k^2 - 4k + 4 = k^2 + 9\)
Cancel \(k^2\) and \(9\) from both sides:
\(4 - 4k = 0\)
\(4k = 4 \implies k = 1\)
In simple words: Equate the squares of the distances from the given point to both target points and solve for k.

Exam Tip: Squaring both sides eliminates square roots and simplifies the quadratic terms.

Question 7. Prove that the points (4,5),(7,6),(6,3) and (3,2) are the vertices of a parallelogram. Is it a rectangle?
Answer: Let \(A(4, 5)\), \(B(7, 6)\), \(C(6, 3)\), and \(D(3, 2)\).
1. Find the squares of the lengths of all four sides:
\(AB^2 = (7 - 4)^2 + (6 - 5)^2 = 3^2 + 1^2 = 9 + 1 = 10\)
\(BC^2 = (6 - 7)^2 + (3 - 6)^2 = (-1)^2 + (-3)^2 = 1 + 9 = 10\)
\(CD^2 = (3 - 6)^2 + (2 - 3)^2 = (-3)^2 + (-1)^2 = 9 + 1 = 10\)
\(DA^2 = (4 - 3)^2 + (5 - 2)^2 = 1^2 + 3^2 = 1 + 9 = 10\)
Since opposite sides are equal (\(AB = CD = \sqrt{10}\) and \(BC = DA = \sqrt{10}\), and all four sides are equal), it is a rhombus, hence also a parallelogram.
2. Check the diagonals to test if it is a rectangle:
\(AC^2 = (6 - 4)^2 + (3 - 5)^2 = 2^2 + (-2)^2 = 4 + 4 = 8\)
\(BD^2 = (3 - 7)^2 + (2 - 6)^2 = (-4)^2 + (-4)^2 = 16 + 16 = 32\)
Since the diagonals are not equal (\(AC \neq BD\), as \(8 \neq 32\)), it is not a rectangle (it is a rhombus).
In simple words: A figure with four equal sides is a rhombus. Since its diagonals are unequal, it is not a rectangle.

Exam Tip: To prove a quadrilateral is a parallelogram, showing opposite sides are equal is sufficient. To check if it is a rectangle, check if diagonals are equal.