RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 9 Co-ordinate Geometry Exercise 9.1.

## Rajasthan Board RBSE Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1

Question 1.

Find the co-ordinates of points P, Q, R and S from given figure.

Solution :

Coordinate of P (5, 3)

Coordinate of Q (-4, 6)

Coordinate of R (-3, -2)

Coordinate of S = (1, -5)

Question 2.

Plot the points of the following co-ordinates.

(1, 2), (-1, 3), (-2, -4), (3, -2) (2, 0), (0, 3)

Solution :

In the question for plotting the points draw two axis XOX’ and YOY’. and mark the points P(1, 2), Q(-1, 3), R(-2, -4), S(3, -2), T(2, 0) and L(0, 3)

Question 3.

By taking rectangular coordinate axis plot the points 0 (0,0), P (3, 0) and R(O, 4). If OPQR is rectangle, then find coordinates of Q.

Solution :

To plot the points first of all we draw two perpendicular axis XOX’ and YOY’.

Now mark the given points 0(0, 0), P(3, 0) and R(0, 4) which is according to figure.

∵ OPQR is a rectangle. According to graph point Q is at 3 unit from O on X axis and at 4 unit from O on Y axis. Hence, Q has abscissa 3 and ordinate 4.

So, coordinate of point Q = (3, 4)

Question 4.

Plot the points (-1, 0), (1, 0), (1, 1), (0, 2), (-1, 1). Which figure is obtained, by joining them serially?

Solution :

Let points are P(1, 0), Q(1, 1), R(0, 2), S(-1, 1), T(-1, 0) which is shown in figure.

The given figure is pentagon.

Question 5.

Draw quadrilateral, if its vertices are following:

(i) (1, 1), (2, 4), (8, 4) and (10, 1)

(ii) (-2, -2), (-4, 2), (-6, -2) and (-4, -6)

Also, mention type of obtained quadrilateral.

Solution :

(i) First of all use draw two axis XOX’ and YOY’ let the vertices of quadrilateral are P(1, 1), Q(2, 4), R(8, 4) and S (10, 1).

Draw these points on coordinate axis as follows :

It is clear from figure, sides PS and QR are parallel and side PQ and RS are unequal. Hence, given quadrilateral is a trapezium.

(ii) First we draw two axis XOX’ and YOY’. Let vertices of quadrilateral are P(-2, -2), Q(-4, 2), R(-6, -2) and S(-4, -6). Plot these points at coordinate axis from figure it is clear that

PQ = QR = RS = SP

So, given quadrilateral is a rhombus.

Question 6.

Find the distance between the followings points :

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

(ii)(-1, -1) and (8, -2)

(iii) (at_{1}^{2}, 2at_{1})and (at_{2}^{2}, 2at_{2})

Solution :

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

Here x_{1} = -6, y_{1} = 7, x_{2} = -1, y_{2} = 5

We know that distance between two points.

(ii) (-1, -1) and (8, -2)

Here x_{1} = -1, y_{1} = -1, x_{2} = 8, y_{2} = -2

∵ Distance between two points

(iii) (at_{1}^{2}, 2at_{1}) and (at_{2}^{2}, 2at_{2})

Here x_{1} = at_{1}^{2}, y_{1} = 2at_{1}, x_{2} = at_{2}^{2}, y_{2} = 2at_{2}

∵ Distance between two points

Question 7.

Prove that the points (2, -2,) and (5, 2) are vertices of a right angled triangle.

Solution :

Let A(2, -2), B(-2, 1) and C(5, 2) are the vertices of a triangle.

So, Converse of Pythagoras theorem given points are the vertex of right angled triangle.

Question 8.

Prove hat the points (1, -2), (3, 9), (1, 2) and (-1, 0) are the vertices of a square.

Solution :

Let A(1, -2), B(3, 0), C(1, 2) and D(-1, 0) are the vertices of a square.

Hence, sides AB, BC,CD and DA are equal and Diagonal AC = Diagonal BD.

Hence, ABCD is a square and given points are the vertices of a square.

Question 9.

Prove that points (a, a), (a, -a) and (-√3a, √3a) are vertices of an equilateral triangle.

Solution :

Let P(a, a), Q(-a, -a) and R (-√3a, √3a) arc the vertices of a triangle.

∵ PQ = QR = RP = 2√2a

Hence, these points are the vertices of a equilateral triangle.

Question 10.

Prove that points (1, 1), (-2, 7) and (3, -3) are collinear.

Solution :

Let the given points A( 1, 1), B(-2, 7) and C(3, -3) are lie in a line then

Hence, these points are collinear.

Question 11.

Find that point on x axis which is equidistant from points (-2, -5) and (2, -3).

Solution :

We know that on x-axis y coordinate is zero of a point. Let P(x, 0) is at equal distance from points A(-2, -5) and B (2, -3).

∴ PA = PB

Hence, required point (-2, 0)

Question 12.

Find the point on y axis which is equidistant from point (-5, -2) and (3, 2).

Solution :

We know that on y-axis the x d coordinate of any point is zero. Hence on y-axis let any point P(0, y) is equal distance from points A(-5, -2) and B(3, 2)

∴ PA = PB

Hence, required point (0, -2)

Question 13.

If points (3, K) and (K, 5) are equidistant from a point (0, 2), then find the value of K.

Solution :

Let P(0, 2), A(3, K) and B(K, 5).

According to question point P(0, 2) is equal distance from points A(3, K) and B(K, 5)

squaring both sides

K^{2} – 4K + 13 = K^{2} + 9

4K = 13 – 9

4K = 4

K = 1

So, K = 1

Question 14.

IF co-ordinates of P and Q are (a cos θ, b sin θ) and (-a sin θ, b cos θ) respectively, then show that

OP^{2} + OQ^{2} = a^{2} + b^{2}, where O is origin.

Solution :

Let coordinate of P and Q are P(a cos θ, b sin θ) and Q(-a sin θ, b cos θ) Then distance between the points O(0, 0) and P(a cos θ, b sin θ)

Question 15.

If (0, 0) and (3√3) are two vertices of an equilateral triangle, then find third vertex.

Solution :

Let the coordinate of third vertex C of equilateral triangle is (x, y)

According to question, two vertices of equilateral triangle is A(0, 0) and B(3√3)

Squaring both sides

(2√3y)^{2} = (12 – 6x)^{2}

12y^{2} = 144 + 36x^{2}– 144x

⇒ 12(12 – x^{2}) = 144 + 36x^{2} – 144x [x^{2} + y^{2} = 12]

⇒ 144 – 12x^{2} = 144 + 36x^{2} – 144x

⇒ – 12x^{2} = 36x^{2} – 144

⇒ 36x^{2} + 12x^{2} – 144x = 0

⇒ 48x^{2} – 144x = 0

⇒ 48x(x – 3) = 0

⇒ x = 0 or x – 3 = 0

⇒ x = 0 or x = 3

Hence, put x = 0 in equation (v)

x^{2} + y^{2} = 12

0 + y^{2} = 12

y^{2} = 12

y = ± 2√3

Hence, x = 0, y = ± 2√3

Put x = 3 in equation (v)

x^{2} + y^{2} = 12

(3)^{2} + y^{2} = 12

y^{2} = 12 – 9

y = ± √3

From x = 3, y = ± √3

Hence, coordinate of third vertex is (0, 2√3), (0, -2√3), (3, √3) and (3, -√3)

(3, √3) is given. Hence coordinate of third vertex is (0,+ 2√3) or (0, – 2.√3) or (3, -√3)

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