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RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise

April 24, 2019 by Fazal Leave a Comment

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

Rajasthan Board RBSE Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise

Multiple Choice Questions [1 to 10]
Question 1.
Distance of (RBSESolutions.com) point (3, 4) from y-axis will be :
(A) 1
(B) 4
(C) 2
(D) 3
Solution :
Distance of point (3, 4) from y-axis = 3 unit.
Hence, correct choice is (D).

Question 2.
Distance of point (5, -2) from x-axis will be
(A) 5
(B) 2
(C) 3
(D) 4
Solution :
Distance of point (5, -2) from x-axis = 2 unit
So, correct choice is (B).

RBSE Solutions

Question 3.
Distance between (RBSESolutions.com) points (0, 3) and (-2, 0) will be :
(A) \(\sqrt { 14 }\)
(B) \(\sqrt { 15 }\)
(C) \(\sqrt { 13 }\)
(D) \(\sqrt { 5 }\)
Solution :
Let A(0, 3) and B(-2, 0) are two point.
So distance between them
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 1
Hence, correct choice is (C)

Question 4.
Triangle having vertices (-2, 1), (2, -2) and (5, 2) is :
(A) Right angle
(B) Equilateral
(C) Isosceles
(D) None of these
Solution :
Let the vertices of given triangle is A(-2, 1), B(2, -2) and C(5, 2), then by distance formula
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 2
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 3
So given triangle is a right angled triangle.
So, correct choice is (A)

Question 5.
Quadrilateral having (RBSESolutions.com) vertices (-1, 1), (0, -3), (5, 2) and (4, 6) will be :
(A) (1, 2)
(B) (2, 1)
(C) (2, 2)
(D) (1, 1)
Solution :
We draw the given points on rectangular co-ordinate axis, we get parallelogram.
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 4
So, correct choice is (D)

Question 6.
Point equidistant from (0, 0), (2, 0) and (0, 2) is :
(A) (1, 2)
(B) (2, 1)
(C) (2, 2)
(D) (1, 1)
Solution :
Let P(x,y) is a equidistant (RBSESolutions.com) from the points A(0, 0), B(2, 0) and C (0, 2). So.
PA = PB = PC
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 5
∵ PA = PB
Squaring both sides.
⇒ PA2 = PB2
⇒ x2 + y2 = (x – 2)2 + y2
⇒ x2 = x2 + 4 – 4x
⇒ 4x = 4
⇒ x = 1
Again PA = PC
Squaring both sides.
⇒ PA2 = PC2
⇒ x2 + y2 = x2 + (y – 2)2
⇒ y2 = y2 + 4 – 4y
⇒ 4y = 4
⇒ y = 1
Here, required point is (1, 1)
Hence, correct choice is (D)

Question 7.
P divides internally the line segment (RBSESolutions.com) which joins the points (5, 0) and (0, 4) in the ratio of 2 : 3 internally. Co-ordinates of point P is :
(A) \(\left( 3,\frac { 8 }{ 5 } \right)\)
(B) \(\left( 2,\frac { 8 }{ 5 } \right)\)
(C) \(\left( \frac { 5 }{ 2 } ,\frac { 3 }{ 4 } \right)\)
(D) \(\left( 2,\frac { 12 }{ 5 } \right)\)
Solution :
Let point P(x, y) divides the line segment joining the points A(5, 0) and B(0, 4) internally in the ratio 2 : 3.
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 6
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 7
Hence, co-ordinate of P is \(\left( 3,\frac { 8 }{ 5 } \right)\)
Hence, correct choice is (A)

RBSE Solutions

Question 8.
If points (1, 2), (-1, x) and (2, 3) are collinear, (RBSESolutions.com) then x will be :
(A) 2
(B) 0
(C) -1
(D) 1
Solution :
Let the points A(1, 2), B(1, x) and C(2, 3) are collinear then area of triangle made by these points will be zero.
⇒ \(\frac { 1 }{ 2 }\)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
⇒ \(\frac { 1 }{ 2 }\)[(x – 3) + (-1)(3 – 2) + 2(2 – x)] = 0
⇒ x – 3 – 1 + 2(2 – x) = 0
⇒ x – 3 – 1 + 4 – 2x = 0
⇒ -x – 4 + 4 = 0
⇒ x = 0
So, correct choice is (B)

Question 9.
If distance between point (3, a) and (4, 1) is \(\sqrt { 10 }\), then a will be :
(A) 3, -1
(B) 2, -2
(C) 4, -2
(D) 5, -3
Solution :
According to question, distance between the (RBSESolutions.com) points A(3, a) and B(4, 1) is \(\sqrt { 10 }\)
i.e. AB = \(\sqrt { 10 }\)
⇒ \(\sqrt { { \left( 4-3 \right) }^{ 2 }+{ \left( 1-a \right) }^{ 2 } } =\sqrt { 10 } \)
⇒ 1 + (1 – a)2 = 10
⇒ (1 – a)2 = 10 – 1 = 9
⇒ a – 1 = ±3
taking +ve sign
a – 1 = 3
a = 3 + 1
a = 4
taking -ve sign
a – 1 = -3
a = 3 + 1
a = -2
So, a = 4, -2
Hence, correct choice in (C).

Question 10.
If point (x, y) is at equidistant from point (2, 1) and (1, -2), then choose the true statement of the following :
(A) x + 3y = 0
(B) 3x + y = 0
(C) x + 2y = 0
(D) 2x + 3y = 0
Solution :
Let P(x,y) is a equal distant from the (RBSESolutions.com) points A(2, 1) and B (1, -2)
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 8
PA = \(\sqrt { { \left( x-2 \right) }^{ 2 }+{ \left( y-1 \right) }^{ 2 } } \)
PB = \(\sqrt { { \left( x-1 \right) }^{ 2 }+{ \left( y+2 \right) }^{ 2 } } \)
∵ PA = PB
Squaring both sides
⇒ PA2 = PB2
(x – 2)2 + (y – 1)2 = (x – 1)2 + (y + 2)2
⇒ x2 + 4 – 4x + y2 + 1 – 2y = x2 + 1 – 2x + y2 + 4 + 4y
⇒ -4x – 2y + 5 = -2x + 4y + 5
⇒ – 4x + 2x – 2y – 4y + 5 – 5 = 0
⇒ -2x – 6y = 0
⇒ -2(x + 3y) = 0
⇒ x + 3y = 0
Hence, correct choice is (A).

Question 11.
Find the type of quadrilateral, If its (RBSESolutions.com) vertices are (1, 4), (-5, 4), (-5, -3) and (1, -3).
Solution :
Let the vertices of quadrilateral are A( 1, 4), B(-5, 4) C(-5, -3) and D(1, -3) respectively. Then
Hence, AB = CD, and BC = DA and Diagonal BD = Diagonal AC
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 9
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 10
Hence, these points are the (RBSESolutions.com) vertices of rectangle.

RBSE Solutions

Question 12.
which shape will be formed on joining (-2, 0), (2, 0), (2, 2), (0, 4), (-2, 2) in the given order?
Solution :
First of all we draw co-ordinate axis XOX’ and YOY’ and mark the points A(-2, 0), B(2, 0), C(2, 2) D(0, 4) and E(-2, 2) then we get pentagon.
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 11

Question 13.
Find the ratio ¡n which point (3, 4) divides the (RBSESolutions.com) line segment which joins points (1, 2) and (6, 7).
Solution :
Let point P(3, 4) divides the line segment joining the points A(1, 2) and B(6, 7) internally in the ratio m1 : m2
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 12
⇒ 3(m1 + m2) = 6m1 + m2
⇒ 3m1 + 3m2 = 6m1 + m2
⇒ 3m1 – 6m1 = m2 – 3m2
⇒ -3m1 = -2m2
⇒ \(\frac { { m }_{ 1 } }{ { m }_{ 2 } } =\frac { -2 }{ -3 } \) = \(\frac { 2 }{ 3 }\)
⇒ m1 : m2 = 2 : 3
Hence required ratio 2 : 3

Question 14.
An opposite vertices of any (RBSESolutions.com) square are (5, -4) and (-3, 2), then find the length of diagonal.
Solution :
Let the vertices of square are A(5, -4) and C(-3, 2), then
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 13
Hence, length of the diagonal = 10 unit

Question 15.
If co-ordinate of one end and midpoint of a line (RBSESolutions.com) segment are (4, 0) and (4, 1) respectively, then find the co-ordinate of other end of line segment.
Solution :
Let co-ordinate of side A is (4, 0) and co-ordinate of side B is (x, y) of line segment AB. The co-ordinate of mid point P is (4, 1)
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 14
Hence, co-ordinate of point B is (4, 2)

Question 16.
Find the distance between the point (1, 2) from (RBSESolutions.com) mid point of line segment which joint the points (6, 8) and (2, 4).
Solution :
Let side of line segment AB is A(6, 8) and B(2, 4)
Co-ordinate of mid point P of AB
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 15
Now distance between the points P(4, 6) and C(1, 2)
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 16
Hence, required distance = 5 unit

RBSE Solutions

Question 17.
If in any plane there are four (RBSESolutions.com) points P(2, -1), Q(3, 4), R(-2, 3) and S(-3, -2), then prove that PQRS is not a square but a rhombus.
Solution :
Let four points P(2, -1), Q(3, 4), R(-2, 3) and S(-3, -2) are in a plane.
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 17
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 18
∵ PQ = QR = RS = SP = \(\sqrt { 26 }\)
And diagonal PR ≠ diagonal SQ
Since diagonals are also equal in a square (RBSESolutions.com) but here diagonals are not equal. Hence PQRS is not square but a rhombus.

Question 18.
Prove that mid point (C) of hypotaneous ¡n a right angled triangle AOB is situated at equal distance from vertices O, A and B of triangle.
Solution :
Let in right angled triangle AOB, the vertices 0(0, 0), A(a, 0) and B(0, b) are shown in following.
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 19
Now co-ordinate of mid (RBSESolutions.com) point C of hypotaneous
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 20
Clearly : OC = CA = CB
Hence, the mid point C of hypotaneous in a (RBSESolutions.com) right angled triangle AOB is situated at equal distance from vertices O, A and B of the triangle.

RBSE Solutions

Question 19.
Find the length of median of triangle whose vertices are (1, -1), (0, 4) and (-5, 3) respectively.
Solution :
Let the vertices of triangle are A(1, -1), B(0, 4) and C(-5, 3). Let AP, BQ and CR are medians drawn from vertices A, B and C respectively.
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 21
The co-ordinate of mid point P of side BC.
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 22
Hence, co-ordinate (RBSESolutions.com) of P = \(\left( \frac { -5 }{ 2 } ,\frac { 7 }{ 2 } \right)\)
And co-ordinate of mid point Q of side CA.
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 23
Now co-ordinate of mid point R of side AB
= \(\left( \frac { 1+0 }{ 2 } ,\frac { -1+4 }{ 2 } \right)\)
= \(\left( \frac { 1 }{ 2 } ,\frac { 3 }{ 2 } \right)\)
Hence, the co-ordinate of R = \(\left( \frac { 1 }{ 2 } ,\frac { 3 }{ 2 } \right)\)
∴ Length of median AP = Distance between the points A(1, -1) and point p\(\left( \frac { -5 }{ 2 } ,\frac { 7 }{ 2 } \right)\)
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 24
Length of median BQ = Distance (RBSESolutions.com) between the points B(0, 4) and Q(2, 1)
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 25
Length of median CR = Distance between the points C(-5, 3) and R\(\left( \frac { 1 }{ 2 } ,\frac { 3 }{ 2 } \right)\)
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 26
Hence length of median are \(\frac { \sqrt { 130 } }{ 2 }\), \(\sqrt { 13 }\) and \(\frac { \sqrt { 130 } }{ 2 }\) respectively.

Question 20.
Prove that mid point of a line segment which (RBSESolutions.com) joins the points (5, 7) and (3, 9) is the same as mid point of line segment which joins the points (8, 6) and (3, 10).
Solution :
Coordinate of mid point joining the live segment of points (5, 7) and (3, 9) is
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 27
Again, Coordinate of mid point of line segment joining the points (8, 6) and (0, 10)
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 28
Clearly mid point of line segment which joins the (RBSESolutions.com) points (5, 7) and (3, 9) is the same as mid point of line segment which joins the points (5, 7) and (3, 9).

RBSE Solutions

Question 21.
If mid points of sides of a triangle is (1, 2), (0, -1) and (2, -1), then find its vertices.
Solution :
Let P(1, 2), Q(0, -1) and R(2, -1) are mid point of sides of given triangle and coordinate of vertices of triangle are A(x1, y1), B(x2, y2) and C(x3, y3) respectively.
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 29
Since, co-ordinate of mid (RBSESolutions.com) point P is (1, 2) of A(x1, y1) and B(x2, y2) then
1 = \(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } \) and 2 = \(\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \)
⇒ x1 + x2 = 2 …..(i)
and y1 + y2 = 4 …(ii)
Since Q(0, -1) is the mid point of B(x2, y2) and C(x3, y3)
So \(\frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } \) = 0, \(\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } \) = -1,
⇒ x2 + x3 = 0 …(iii)
and y2 + y3 = -2 …(iv)
Since R(2, -1) is the mid point of A(x1, y1) and C(x3, y3)
So \(\frac { { x }_{ 1 }+{ x }_{ 3 } }{ 2 } \) = 2, \(\frac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 } \) = -1,
⇒ x1 + x3 = 4 …(v)
and y1 + y3 = -2 …(vi)
adding equation (i), (iii) and (v)
2x1 + 2x2 + 2x3 = 6
= x1 + x2 + x3 = 3 …(vii)
Adding equation (ii), (iv) and (vi)
2y1 + 2y2 + 2y3 = 0
y1 + y2 + y3 = 0 …..(viii)
Put the value of equation (i) in equation (vii)
x1 + x2 + x3 = 3
2 + x3 = 3
x3 = 3 – 2 = 1
Put the value of equation (ii) in equation (viii)
y1 + y2 + y3 =0
4 + y3 = 0
y3 = -4
Hence, co-ordinate (RBSESolutions.com) of point C (x3, y3) = (1, -4)
Subtract equation (iii) from equation (vii)
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 30
Hence, co-ordinate of point A(x1, y1) = (3, 2)
Again subtract equation (v) from equation (vii)
RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Miscellaneous Exercise 31
Hence, co-ordinate (-1, 2)
Hence, the vertices of (RBSESolutions.com) triangle are (1, -4), (3, 2) and (-1, 2) Respectively.

RBSE Solutions

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

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