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).

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

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

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.

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

∵ PA = PB

Squaring both sides.

⇒ PA^{2} = PB^{2}

⇒ x^{2} + y^{2} = (x – 2)^{2} + y^{2}

⇒ x^{2} = x^{2} + 4 – 4x

⇒ 4x = 4

⇒ x = 1

Again PA = PC

Squaring both sides.

⇒ PA^{2} = PC^{2}

⇒ x^{2} + y^{2} = x^{2} + (y – 2)^{2}

⇒ y^{2} = y^{2} + 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.

Hence, co-ordinate of P is \(\left( 3,\frac { 8 }{ 5 } \right)\)

Hence, correct choice is (A)

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.

⇒ \(\frac { 1 }{ 2 }\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

⇒ \(\frac { 1 }{ 2 }\)[(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)

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

⇒ PA^{2} = PB^{2}

(x – 2)^{2} + (y – 1)^{2} = (x – 1)^{2} + (y + 2)^{2}

⇒ x^{2} + 4 – 4x + y^{2} + 1 – 2y = x^{2} + 1 – 2x + y^{2} + 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

Hence, these points are the (RBSESolutions.com) vertices of rectangle.

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.

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 m_{1} : m_{2}

⇒ 3(m_{1} + m_{2}) = 6m_{1} + m_{2}

⇒ 3m_{1} + 3m_{2} = 6m_{1} + m_{2}

⇒ 3m_{1} – 6m_{1} = m_{2} – 3m_{2}

⇒ -3m_{1} = -2m_{2}

⇒ \(\frac { { m }_{ 1 } }{ { m }_{ 2 } } =\frac { -2 }{ -3 } \) = \(\frac { 2 }{ 3 }\)

⇒ m_{1} : m_{2} = 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

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)

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

Now distance between the points P(4, 6) and C(1, 2)

Hence, required distance = 5 unit

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.

∵ 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.

Now co-ordinate of mid (RBSESolutions.com) point C of hypotaneous

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.

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.

The co-ordinate of mid point P of side BC.

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.

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\(\left( \frac { -5 }{ 2 } ,\frac { 7 }{ 2 } \right)\)

Length of median BQ = Distance (RBSESolutions.com) between the points B(0, 4) and Q(2, 1)

Length of median CR = Distance between the points C(-5, 3) and R\(\left( \frac { 1 }{ 2 } ,\frac { 3 }{ 2 } \right)\)

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

Again, Coordinate of mid point of line segment joining the points (8, 6) and (0, 10)

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).

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(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) respectively.

Since, co-ordinate of mid (RBSESolutions.com) point P is (1, 2) of A(x_{1}, y_{1}) and B(x_{2}, y_{2}) then

1 = \(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } \) and 2 = \(\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \)

⇒ x_{1} + x_{2} = 2 …..(i)

and y_{1} + y_{2} = 4 …(ii)

Since Q(0, -1) is the mid point of B(x_{2}, y_{2}) and C(x_{3}, y_{3})

So \(\frac { { x }_{ 2 }+{ x }_{ 3 } }{ 2 } \) = 0, \(\frac { { y }_{ 2 }+{ y }_{ 3 } }{ 2 } \) = -1,

⇒ x_{2} + x_{3} = 0 …(iii)

and y_{2} + y_{3} = -2 …(iv)

Since R(2, -1) is the mid point of A(x_{1}, y_{1}) and C(x_{3}, y_{3})

So \(\frac { { x }_{ 1 }+{ x }_{ 3 } }{ 2 } \) = 2, \(\frac { { y }_{ 1 }+{ y }_{ 3 } }{ 2 } \) = -1,

⇒ x_{1} + x_{3} = 4 …(v)

and y_{1} + y_{3} = -2 …(vi)

adding equation (i), (iii) and (v)

2x_{1} + 2x_{2} + 2x_{3} = 6

= x_{1} + x_{2} + x_{3} = 3 …(vii)

Adding equation (ii), (iv) and (vi)

2y_{1} + 2y_{2} + 2y_{3} = 0

y_{1} + y_{2} + y_{3} = 0 …..(viii)

Put the value of equation (i) in equation (vii)

x_{1} + x_{2} + x_{3} = 3

2 + x_{3} = 3

x_{3} = 3 – 2 = 1

Put the value of equation (ii) in equation (viii)

y_{1} + y_{2} + y_{3} =0

4 + y_{3} = 0

y_{3} = -4

Hence, co-ordinate (RBSESolutions.com) of point C (x_{3}, y_{3}) = (1, -4)

Subtract equation (iii) from equation (vii)

Hence, co-ordinate of point A(x_{1}, y_{1}) = (3, 2)

Again subtract equation (v) from equation (vii)

Hence, co-ordinate (-1, 2)

Hence, the vertices of (RBSESolutions.com) triangle are (1, -4), (3, 2) and (-1, 2) Respectively.

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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