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]
RBSE Class 10 Maths Chapter 9 Miscellaneous 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).
RBSE Solutions For Class 10 Maths Chapter 9 Miscellaneous 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 For Class 10 Maths Chapter 9 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)
RBSE Class 10 Maths Solutions Chapter 9 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)
RBSE Class 10 Maths Chapter 9 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)
Chapter 9 Class 10 Maths RBSE 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.
⇒ 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)
RBSE Solutions For Class 10 Maths Chapter 9.1 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)
Chapter 9 Maths Class 10 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)
Ch 9 Class 10 Maths RBSE 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).
Class 10 RBSE Maths Solution 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
⇒ 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).
Class 10 RBSE Maths Solution Ch 9 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.
Class 10 RBSE Maths Chapter 9 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.
Ch 9 Maths Class 10 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
⇒ 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
Chapter 9 Class 10 Maths 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
RBSE Class 10 Chapter 9 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)
RBSE Solutions 10 Maths 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
Www.RBSEsolutions.Com Class 10 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.
Chapter 9 Maths Class 10 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.
RBSE Solutions For Class 10 Maths Chapter 5 Miscellaneous 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.
RBSE Solutions For Class 10 Maths Chapter 9.1 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).
Ncert Solutions For Class 10 Maths Chapter 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(x1, y1), B(x2, y2) and C(x3, y3) respectively.
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)
Hence, co-ordinate of point A(x1, y1) = (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.
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