## RD Sharma Class 10 Solutions Chapter 14 Coordinate Geometry Ex 14.2

### RD Sharma Class 10 Solutions Co-Ordinate Geometry Exercise 14.2

Question 1.

Find the distance between the following pair of points :

(i) (-6, 7) and (-1, -5)

(ii) (a + b, b + c) and (a – b, c – b)

(iii) (a sin α, -b cos α) and (-a cos α, -b sin α)

(iv) (a, 0) and (0, b)

Solution:

Question 2.

Find the value of a when the distance between the points (3, a) and (4, 1) is √10

Solution:

Question 3.

If the points (2, 1) and (1, -2) are equidistant from the point (x, y), show that x + 3y = 0.

Solution:

Question 4.

Find the values of x, y if the distances of the point (x, y) from (-3, 0) as well as from (3, 0) are 4.

Solution:

Distance between (x, y) and (-3, 0) is

Question 5.

The length of a line segment is of 10 units and the coordinates of one end-point are (2, -3). If the abscissa of the other end is 10, find the ordinate of the other end.

Solution:

Let the ordinate of other end by y, then The distance between (2, -3) and (10, y) is

Question 6.

Show that the points (-4, -1), (-2, -4), (4, 0) and (2, 3) are the vertices points of a rectangle. (C.B.S.E. 2006C)

Solution:

AB = CD and AD = BC

and diagonal AC = BD

ABCD is a rectangle

Question 7.

Show that the points A (1, -2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram.

Solution:

Question 8.

Prove that the points A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4) are the vertices of a square. [NCERT]

Solution:

Vertices A (1, 7), B (4, 2), C (-1,-1), D (-4, 4)

If these are the vertices of a square, then its diagonals and sides are equal

Question 9.

Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right-angled isosceles triangle. (C.B.S.E. 2006C)

Solution:

Question 10.

Prove that (2, -2), (-2, 1) and (5, 2) are the vertices of a right angled triangle. Find the area of the triangle and the length of the hypotenuse.

Solution:

Question 11.

Prove that the points (2a, 4a), (2a, 6a) and (2a + √3 a , 5a) are the vertices of an equilateral triangle.

Solution:

Question 12.

Prove that the points (2, 3), (-4, -6) and (1, \(\frac { 3 }{ 2 }\) )do not form a triangle.

Solution:

Question 13.

The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of ∆ABC. [NCERT Exemplar]

Solution:

Given that, the points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a ∆ABC right angled at B.

By Pythagoras theorem, AC² = AB² + BC² ………(i)

Now, by distance formula,

Question 14.

Show that the quadrilateral whose vertices are (2, -1), (3, 4), (-2, 3) and (-3, -2) is a rhombus.

Solution:

Question 15.

Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3.

Solution:

Two vertices of an isosceles ∆ABC are A (2, 0) and B (2, 5). Let co-ordinates of third vertex C be (x, y)

Question 16.

Which point on x-axis is equidistant from (5, 9) and (-4, 6) ?

Solution:

Let co-ordinates of two points are A (5, 9), B (-4, 6)

The required point is on x-axis

Its ordinates or y-co-ordinates will be 0

Let the co-ordinates of the point C be (x, 0)

AC = CB

Question 17.

Prove that the point (-2, 5), (0, 1) and (2, -3) are collinear.

Solution:

Now AB + BC = 2√5 +2√5

and CA = 4√5

AB + BC = CA

A, B and C are collinear

Question 18.

The co-ordinates of the point P are (-3,2). Find the co-ordinates of the point Q which lies on the line joining P and origin such that OP = OQ.

Solution:

Co-ordinates of P are (-3, 2) and origin O are (0, 0)

Let co-ordinates of Q be (x, y)

O is the mid point of PQ

= 9 + 4 = (±3)² + (±2)²

The point will be in fourth quadrant

Its y-coordinates will be negative

and x-coordinates will be positive

Now comparing the equation

x² = (±3)² => x = ±3

y² = (±2)² => y = ±2

x = 3, y = -2

Co-ordinates of the point Q are (3, -2)

Question 19.

Which point on y-axis is equidistant from (2, 3) and (-4, 1) ?

Solution:

The required point lies on y-axis

Its abscissa will be zero

Let the point be C (0, y) and A (2, 3), B (-4, 1)

Now,

Question 20.

The three vertices of a parallelogram are (3, 4), (3, 8) and (9, 8). Find the fourth vertex.

Solution:

Let ABCD be a parallelogram and vertices will be A (3, 4), B (3, 8), C (9, 8)

Question 21.

Find a point which is equidistant from the point A (-5, 4) and B (-1, 6). How many such points are there? [NCERT Exemplar]

Solution:

Let P (h, k) be the point which is equidistant from the points A (-5, 4) and B (-1, 6).

PA = PB

So, the mid-point of AB satisfy the Eq. (i).

Hence, infinite number of points, in fact all points which are solution of the equation 2h + k + 1 = 0, are equidistant from the point A and B.

Replacing h, k, by x, y in above equation, we have 2x + y + 1 = 0

Question 22.

The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, -9) and has diameter 10√2 units. [NCERT Exemplar]

Solution:

By given condition,

Distance between the centre C (2a, a-1) and the point P (11, -9), which lie on the circle = Radius of circle

Question 23.

Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office. What is the extra distance travelled by Ayush in reaching the office? (Assume that all distance covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in kilometers. [NCERT Exemplar]

Solution:

Question 24.

Find the value of k, if the point P (0, 2) is equidistant from (3, k) and (k, 5).

Solution:

Let P (0, 2) is equidistant from A (3, k) and B (k, 5)

PA = PB

=> PA² = PB²

Question 25.

If (-4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the

(i) interior,

(ii) exterior of the triangle. [NCERT Exemplar]

Solution:

Let the third vertex of an equilateral triangle be (x, y).

Let A (-4, 3), B (4,3) and C (x, y).

We know that, in equilateral triangle the angle between two adjacent side is 60 and all three sides are equal.

But given that, the origin lies in the interior of the ∆ABC and the x-coordinate of third vertex is zero.

Then, y-coordinate of third vertex should be negative.

Hence, the require coordinate of third vertex,

C = (0, 3 – 4√3). [C ≠ (0, 3 + 4√3)]

Question 26.

Show that the points (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus. Find the area of this rhombus.

Solution:

Let the co-ordinates of the vertices A, B, C and D of a rhombus are A (-3, 2), B (-5, -5), C (2, -3) and D (4, 4)

Question 27.

Find the coordinates of the circumcentre of the triangle whose vertices are (3, 0), (-1, -6) and (4, -1). Also, find its circumradius.

Solution:

Let ABC is a triangle whose vertices are A (3, 0), B (-1, -6) and C (4, -1)

Question 28.

Find a point on the x-axis which is equidistant from the points (7, 6) and (-3, 4). [CBSE 2005]

Solution:

The required point is on x-axis

Its ordinate will be O

Let the co-ordinates of the required point P (x, 0)

Let the point P is equidistant from the points A (7, 6) and B (-3, 4)

Question 29.

(i) Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square. [CBSE 2004]

(ii) Prove that the points A (2, 3), B (-2, 2), C (-1, -2) and D (3, -1) are the vertices of a square ABCD. [CBSE 2013]

(iii) Name the type of triangle PQR formed by the point P(√2 , √2), Q(- √2, – √2) and R (-√6 , √6 ). [NCERT Exemplar]

Solution:

Question 30.

Find the point on x-axis which is equidistant from the points (-2, 5) and (2, -3). [CBSE 2004]

Solution:

The point P lies on x-axis

The ordinates of P will be 0 Let the point P be (x, 0)

Let P is equidistant from A (-2, 5) and B (2, -3)

Question 31.

Find the value of x such that PQ = QR where the co-ordinates of P, Q and R are (6, -1) (1, 3) and (x, 8) respectively. [CBSE 2005]

Solution:

Question 32.

Prove that the points (0, 0), (5, 5) and (-5, 5) are the vertices of a right isosceles triangle. [CBSE 2005]

Solution:

Let the vertices of a triangle be A (0, 0), B (5, 5) and C (-5, 5)

Question 33.

If the points P (x, y) is equidistant from the points A (5, 1) and B (1,5), prove that x = y. [CBSE 2005]

Solution:

Question 34.

If Q (0, 1) is equidistant from P (5, -3) and R (x, 6) find the values of x. Also find the distances QR and PR. [NCERT]

Solution:

Q (0, 1) is equidistant from P (5, -3) and R (x, 6)

PQ = RQ

Question 35.

Find the values ofy for which the distance between the points P (2, -3) and Q (10, y) is 10 units. [NCERT]

Solution:

Distance between P (2, -3) and Q (10, y) = 10

Question 36.

If the point P (k – 1, 2) is equidistant from the points A (3, k) and B (k, 5), find the values of k. [CBSE 2014]

Solution:

Point P (k – 1, 2) is equidistant from A (3, k) and B (k, 5)

PA= PB

Question 37.

If the point A (0, 2) is equidistant from the point B (3, p) and C (p, 5), find p. Also, find the length of AB. [CBSE 2014]

Solution:

Point A (0, 2) is equidistant from B (3, p) and C (p, 5)

AB = AC

Question 38.

Name the quadrilateral formed, if any, by the following points, and give reasons for your answers :

(i) A (-1, -2), B (1, 0), C (-1, 2), D (-3, 0)

(ii) A (-3, 5), B (3, 1), C (0, 3), D (-1, -4)

(iii) A (4, 5), B (7, 6), C (4, 3), D (1, 2)

Solution:

Question 39.

Find the equation of the perpendicular bisector of the line segment joining points (7, 1) and (3, 5).

Solution:

Let the given points are A (7, 1) and B (3, 5) and mid point be M

Question 40.

Prove that the points (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order, form a rhombus. Also find its area.

Solution:

Question 41.

In the seating arrangement of desks in a classroom three students Rohini, Sandhya and Bina are seated at A (3, 1), B (6, 4) and C (8, 6). Do you think they are seated in a line ?

Solution:

A (3, 1), B (6, 4) and C (8, 6)

Question 42.

Find a point ony-axis which is equidistant from the points (5, -2) and (-3, 2).

Solution:

The point lies on y-axis

Its x = 0

Let the required point be (0, y) and let A (5, -2) and B (-3, 2)

Question 43.

Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4). [NCERT]

Solution:

Question 44.

If a point A (0, 2) is equidistant from the points B (3, p) and C (p, 5), then find the value of p. [CBSE 2012]

Solution:

Point A (0, 2) is equidistant from the points B (3, p) and C (p, 5)

AB = AC

Question 45.

Prove that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle. [CBSE 2013]

Solution:

Let points are A (7, 10), B (-2, 5) and C (3, -4)

Question 46.

If the point P (x, 3) is equidistant from the points A (7, -1) and B (6, 8), find the value of x and And the distance AP. [CBSE 2014]

Solution:

Point P (x, 3) is equidistant from the points A (7, -1) and B (6, 8)

PA = PB

Question 47.

If A (3, y) is equidistant from points P (8, -3) and Q (7, 6), find the value of y and find the distance AQ. [CBSE 2014]

Solution:

Point A (3, y) is equidistant from P (8, -3) and Q (7, 6)

i.e., AP = AQ

Question 48.

If (0, -3) and (0, 3) are the two vertices of an equilateral triangle, find the coordinates of its third vertex. [CBSE 2014]

Solution:

Let A (0, -3) and B (0, 3) are vertices of an equilateral triangle

Let the coordinates of the third vertex be C (x, y)

Question 49.

If the point P (2, 2) is equidistant from the points A (-2, k) and B (-2k, -3), find k. Also, find the length of AP.

Solution:

Point P (2, 2) is equidistant from the points A (-2, k) and B (-2k, -3)

AP = BP

Now,

Question 50.

Show that ∆ABC, where A (-2, 0), B (2, 0), C (0, 2) and ∆PQR, where P (-4, 0), Q (4, 0), R (0, 4) are similar.

Solution:

Question 51.

An equilateral triangle has two vertices at the points (3, 4), and (-2, 3). Find the co-ordinates of the third vertex.

Solution:

Let two vertices of an equilateral triangle are A (3,4), and B (-2,3) and let the third vertex be C (x, y)

Question 52.

Find the circumcentre of the triangle whose vertices are (-2, -3), (-1, 0), (7, -6).

Solution:

Question 53.

Find the angle subtended at the origin by the line segment whose end points are (0, 100) and (10, 0).

Solution:

Let co-ordinates of the end points of a line segment are A (0, 100), B (10, 0) and origin is O (0, 0)

Abscissa of A is 0

It lies on y-axis

Similarly, ordinates of B is 0

It lies on x-axis

But axes intersect each other at right angle

AB will subtended 90° at the origin

Angle is 90° or \(\frac { \pi }{ 2 }\)

Question 54.

Find the centre of the circle passing through (5, -8), (2, -9) and (2, 1).

Solution:

Question 55.

If two opposite vertices of a square are (5, 4) and (1, -6), find the coordinates of its remaining two vertices.

Solution:

Two opposite points of a square are (5, 4) and (1, -6)

Let ABCD be a square and A (5, 4) and C (1, -6) are the opposite points

Let the co-ordinates of B be (x, y). Join AC

Question 56.

Find the centre of the circle passing through (6, -6), (3, -7) and (3, 3).

Solution:

Let O is the centre of the circle is (x, 7) Join OA, OB and OC

Question 57.

Two opposite vertices of a square are (-1, 2) and (3, 2). Find the co-ordinates of other two vertices.

Solution:

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#### RD Sharma Class 10 Solutions

- Chapter 14 Co-Ordinate Geometry Ex 14.1
- Chapter 14 Co-Ordinate Geometry Ex 14.2
- Chapter 14 Co-Ordinate Geometry Ex 14.3
- Chapter 14 Co-Ordinate Geometry Ex 14.4
- Chapter 14 Co-Ordinate Geometry Ex 14.5