CBSE Class 10 Mathematics Coordinate Geometry Worksheet Set 08

Short Answer Questions-

Question. Show that \( (a, a), (-a, -a) \), and \( (-\sqrt{3}a, \sqrt{3}a) \) are vertices of an equilateral triangle. 
Answer: Verification: Distance between \( (a, a) \) and \( (-a, -a) = \sqrt{(2a)^2 + (2a)^2} = \sqrt{8a^2} \) Distance between \( (-a, -a) \) and \( (-\sqrt{3}a, \sqrt{3}a) = \sqrt{(a-\sqrt{3}a)^2 + (-a-\sqrt{3}a)^2} = \sqrt{a^2+3a^2-2\sqrt{3}a^2 + a^2+3a^2+2\sqrt{3}a^2} = \sqrt{8a^2} \) Distance between \( (a, a) \) and \( (-\sqrt{3}a, \sqrt{3}a) = \sqrt{(a+\sqrt{3}a)^2 + (a-\sqrt{3}a)^2} = \sqrt{8a^2} \) All sides are equal.

Question. Find the points on the x-axis which are at a distance of \( 2\sqrt{5} \) from the point \( (7, -4) \). How many such points are there?
Answer: \( (9, 0), (5, 0) \); 2 points

Question. Find the coordinates of the point on the y-axis which is equidistant from the points \( A(5, 3) \) and \( B(1, -5) \). 
Answer: \( \left(0, \frac{1}{2}\right) \)

Question. Show that \( (1, -1) \) is the centre of the circle circumscribing the triangle whose vertices are \( (4, 3), (-2, 3) \) and \( (6, -1) \).
Answer: Verification by showing the distance from \( (1, -1) \) to each vertex is equal (radius = 5).

Question. If \( R(x, y) \) is a point on the line segment joining the points \( P(a, b) \) and \( Q(b, a) \), then prove that \( x + y = a + b \).
Answer: Verification by collinearity or section formula.

Question. If \( A(-2, 1), B(a, 0), C(4, b) \) and \( D(1, 2) \) are vertices of a parallelogram \( ABCD \), find the values of \( a \) and \( b \). Hence find the lengths of its sides. 
Answer: \( a = 1, b = 1; \sqrt{10} \) units

Question. Point \( A \) lies on the line segment \( XY \) joining \( X(6, -6) \) and \( Y(-4, -1) \) in such a way that \( \frac{XA}{XY} = \frac{2}{5} \). If point \( A \) also lies on the line \( 3x + k(y + 1) = 0 \), find the value of \( k \). 
Answer: \( k = 2 \)

Question. In what ratio does the x-axis divide the line segment joining the points \( (-4, -6) \) and \( (-1, 7) \)? Find the coordinates of the point of division.
Answer: \( 6 : 7; \left(\frac{-34}{13}, 0\right) \)

Question. In what ratio does the point \( P(-4, y) \) divide the line segment joining the point \( A(-6, 10) \) and \( B(3, -8) \)? Hence find the value of \( y \). 
Answer: \( 2 : 7; y = 6 \)

Question. Find the value of \( p \) for which the points \( (-5, 1), (1, p) \) and \( (4, -2) \) are collinear.
Answer: \( p = -1 \)

Question. If the points \( P(-3, 9), Q(a, b) \) and \( R(4, -5) \) are collinear and \( a + b = 1 \), find the values of \( a \) and \( b \). 
Answer: \( a = 2, b = -1 \)

Long Answer Questions: 

Question. Find the centre of a circle passing through the points \( (6, -6), (3, -7) \) and \( (3, 3) \).
Answer: \( (3, -2) \)

Question. The points \( A(x_1, y_1), B(x_2, y_2) \) and \( C(x_3, y_3) \) are the vertices of \( \Delta ABC \).
(i) The median from \( A \) meets \( BC \) at \( D \). Find the coordinates of the point \( D \).
(ii) Find the coordinates of the point \( P \) on \( AD \) such that \( AP : PD = 2 : 1 \).
(iii) Find the coordinates of points \( Q \) and \( R \) on medians \( BE \) and \( CF \) respectively such that \( BQ : QE = 2 : 1 \) and \( CR : RF = 2 : 1 \).
(iv) What are the coordinates of the centroid of the triangle \( ABC \)? 
Answer:
(i) \( \left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right) \)
(ii) \( \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \)
(iii) \( Q\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right), R\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \)
(iv) \( \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \)

Question. The mid-point \( P \) of the line segment joining the points \( A(-10, 4) \) and \( B(-2, 0) \) lies on the line segment joining the points \( C(-9, -4) \) and \( D(-4, y) \). Find the ratio in which \( P \) divides \( CD \). Also find the value of \( y \). 
Answer: \( 3 : 2, y = 6 \)

Question. Komal was asked to plot a point 10 units on the left of the origin and other points 4 units directly above the origin. Which of the following are the two points?
(a) (10, 0) and (0, 4)
(b) (-10, 0) and (0, 4)
(c) (10, 0) and (0, -4)
(d) (-10, 0) and (4, 0)
Answer: (b) (-10, 0) and (0, 4)

Question. The perimeter of a triangle with vertices \( (0, 4), (0, 0) \) and \( (3, 0) \) is
(a) 5 units
(b) 12 units
(c) 11 units
(d) 14 units
Answer: (b) 12 units

Question. If the points \( (1, x), (5, 2) \) and \( (9, 5) \) are collinear then value of \( x \) is
(a) \( \frac{5}{2} \)
(b) \( -\frac{5}{2} \)
(c) -1
(d) 1
Answer: (c) -1

Question. Find the value of 'a' so that the point \( (3, a) \) lies on the line represented by \( 2x - 3y = 5 \). 
Answer: \( a = \frac{1}{3} \)

Question. What is the ratio in which the point \( P\left(\frac{-2}{5}, 6\right) \) divides the line joining \( A(-4, 3) \) and \( B(2, 8) \)?
Answer: \( 3 : 2 \)

Question. What is the distance between the points \( A(5 \cos \theta, 0) \) and \( B(0, 5 \sin \theta) \)?
Answer: 5 units

Question. Examine whether the points \( (1, -1), (-5, 7) \) and \( (2, 5) \) are equidistant from the point \( (-2, 3) \)?
Answer: No

Question. Find the value of \( k \) for which the points \( (-5, 1), (1, k) \) and \( (4, -2) \) are collinear. 
Answer: \( k = -1 \)

Question. Find the ratio in which the line segment joining \( A(1, -5) \) and \( B(-4, 5) \) is divided by the x-axis. Also find the coordinates of the point of division.
Answer: \( 1 : 1; \left(-\frac{3}{2}, 0\right) \)

Question. The two opposite vertices of a square are \( (-1, 2) \) and \( (3, 2) \). Find the coordinates of other two vertices.
Answer: \( (1, 0); (1, 4) \)

Question. The base \( AB \) of two equilateral triangles \( ABC \) and \( ABC' \) with side \( 2a \) lies along the x-axis, such that the mid point of \( AB \) is at the origin. Find coordinates of vertices \( C \) and \( C' \) of the triangles.
Answer: \( C(0, \sqrt{3}a), C'(0, -\sqrt{3}a) \)

Question. Show that the points \( A(1, 0), B(5, 3), C(2, 7) \) and \( D(-2, 4) \) are the vertices of a rhombus.
Answer: Verification: \( AB = \sqrt{(5-1)^2 + (3-0)^2} = 5 \) \( BC = \sqrt{(2-5)^2 + (7-3)^2} = 5 \) \( CD = \sqrt{(-2-2)^2 + (4-7)^2} = 5 \) \( DA = \sqrt{(1+2)^2 + (0-4)^2} = 5 \) Since all sides are equal, it is a rhombus.

Assertion-Reason Questions

The following questions consist of two statements—Assertion(A) and Reason(R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true but R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.

Question. Assertion (A) : The point (0, 4) lies on \( y \)-axis.
Reason (R) : The \( x \) co-ordinate of the point on \( y \)-axis is zero.
Answer: Solution : The \( x \) co-ordinate of the point (0, 4) is zero.
\( \therefore \) Point (0, 4) lies on \( y \)-axis.
So, both A and R are true and R is the correct explanation for A.
Hence, option (a) is correct.

Question. Assertion (A) : The value of \( y \) is 6, for which the distance between the points \( P(2, -3) \) and \( Q(10, y) \) is 10.
Reason (R) : Distance between two given points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is given by \( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
Answer: Solution : \( PQ = 10 \)

\( \implies PQ^2 = 100 \)

\( \implies (10 - 2)^2 + (y + 3)^2 = 100 \)

\( \implies (y + 3)^2 = 100 - 64 = 36 \)

\( \implies y + 3 = \pm 6 \)

\( \implies y = -3 \pm 6 \)

\( \implies y = 3, -9 \)
So, A is false but R is true.
Hence, option (d) is correct.

Question. Assertion (A) : The point \( (-1, 6) \) divides the line segment joining the points \( (-3, 10) \) and \( (6, -8) \) in the ratio \( 2 : 7 \) internally.
Reason (R) : Three points \( A, B \) and \( C \) are collinear if \( AB + BC = AC \)
Answer: Solution : Using section formula, we have
\( -1 = \frac{k \times 6 + 1 \times (-3)}{k + 1} \)

\( \implies -k - 1 = 6k - 3 \)

\( \implies 7k = 2 \)

\( \implies k = \frac{2}{7} \)

\( \implies \) Ratio is \( 2 : 7 \) internally.
Also, if three points \( A, B, C \) satisfy \( AB + BC = AC \). \( A, B \) and \( C \) points are collinear.
So, both A and R are true but R is not the correct explanation for A.
Hence, option (b) is correct.

Question. Assertion (A) : If the centroid of a triangle formed by the points \( (a, b), (b, c) \) and \( (c, a) \) is at origin, Then \( a + b + c = 0 \).
Reason (R) : Centroid of a \( \Delta ABC \) with vertices \( A(x_1, y_1), B(x_2, y_2) \) and \( C(x_3, y_3) \) is given by \( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \).
Answer: Solution : Centroid of a triangle with vertices \( (a, b), (b, c) \) and \( (c, a) \) is \( \left( \frac{a + b + c}{3}, \frac{b + c + a}{3} \right) \).

\( \implies \left( \frac{a + b + c}{3}, \frac{b + c + a}{3} \right) = (0, 0) \)

\( \implies a + b + c = 0 \)
So, both A and R are true and R is the correct explanation for A.
Hence, option (a) is correct.