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RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1

April 22, 2019 by Fazal Leave a Comment

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 (RBSESolutions.com) points P, Q, R and S from given figure.
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 1
Solution :
Coordinate of P (5, 3)
Coordinate of Q (-4, 6)
Coordinate of R (-3, -2)
Coordinate of S = (1, -5)

RBSE Solutions

Question 2.
Plot the points of the (RBSESolutions.com) following co-ordinates.
(1, 2), (-1, 3), (-2, -4), (3, -2) (2, 0), (0, 3)
Solution :
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 2
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 (RBSESolutions.com) 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’.
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 3
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 (RBSESolutions.com) 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 :
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 4
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 (RBSESolutions.com) 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 :
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 5
It is clear from figure, sides PS and QR are parallel (RBSESolutions.com) 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.
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 6

Question 6.
Find the distance between the (RBSESolutions.com) followings points :
(i) (-6, 7) and (-1, -5)
(ii)(-1, -1) and (8, -2)
(iii) (at12, 2at1)and (at22, 2at2)
Solution :
(i) (-6, 7) and (-1, -5)
Here x1 = -6, y1 = 7, x2 = -1, y2 = 5
We know that distance between two points.
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 7
(ii) (-1, -1) and (8, -2)
Here x1 = -1, y1 = -1, x2 = 8, y2 = -2
∵ Distance between two points
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 8
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 9
(iii) (at12, 2at1) and (at22, 2at2)
Here x1 = at12, y1 = 2at1, x2 = at22, y2 = 2at2
∵ Distance between (RBSESolutions.com) two points
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 10

RBSE Solutions

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 (RBSESolutions.com) the vertices of a triangle.
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 11
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 12
So, Converse of Pythagoras theorem (RBSESolutions.com) 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.
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 13
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 14
Hence, sides AB, BC,CD and DA are equal (RBSESolutions.com) 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.
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 15
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 16
∵ PQ = QR = RP = 2√2a
Hence, these points are the vertices (RBSESolutions.com) of a equilateral triangle.

RBSE Solutions

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
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 17
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 18
Hence, these points are collinear.

Question 11.
Find that point on x axis which is (RBSESolutions.com) 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
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 19
Hence, required point (-2, 0)

Question 12.
Find the point on y axis which is (RBSESolutions.com) 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
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 20
Hence, required point (0, -2)

Question 13.
If points (3, K) and (K, 5) are (RBSESolutions.com) 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)
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 21
squaring both sides
K2 – 4K + 13 = K2 + 9
4K = 13 – 9
4K = 4
K = 1
So, K = 1

RBSE Solutions

Question 14.
IF co-ordinates of P and Q are (a cos θ, b sin θ) and (-a sin θ, b cos θ) respectively, (RBSESolutions.com) then show that
OP2 + OQ2 = a2 + b2, 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)
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 22

Question 15.
If (0, 0) and (3√3) are two vertices of an (RBSESolutions.com) 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)
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 23
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 24
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 25
Squaring both (RBSESolutions.com) sides
(2√3y)2 = (12 – 6x)2
12y2 = 144 + 36x2– 144x
⇒ 12(12 – x2) = 144 + 36x2 – 144x [x2 + y2 = 12]
⇒ 144 – 12x2 = 144 + 36x2 – 144x
⇒ – 12x2 = 36x2 – 144
⇒ 36x2 + 12x2 – 144x = 0
⇒ 48x2 – 144x = 0
⇒ 48x(x – 3) = 0
⇒ x = 0 or x – 3 = 0
⇒ x = 0 or x = 3
Hence, put x = 0 in (RBSESolutions.com) equation (v)
x2 + y2 = 12
0 + y2 = 12
y2 = 12
y = ± 2√3
Hence, x = 0, y = ± 2√3
Put x = 3 in equation (v)
x2 + y2 = 12
(3)2 + y2 = 12
y2 = 12 – 9
y = ± √3
From x = 3, y = ± √3
Hence, coordinate of third (RBSESolutions.com) 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)

RBSE Solutions

We hope the given RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 9 Co-ordinate Geometry Exercise 9.1, drop a comment below and we will get back to you at the earliest.

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