Class 10 Maths RD Sharma Solutions

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.1
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

Mark the correct alternative in each of the following :
Question 1.
The distance between the points (cosθ, sinθ) and (sinθ, -cosθ) is
(a) √3
(b) √2
(c) 2
(d) 1
Solution:
(b) Distance between (cosθ, sinθ) and (sinθ, -cosθ)

Question 2.
The distance between the points (a cos 25°, 0) and (0, a cos 65°) is
(a) a
(b) 2a
(c) 3a
(d) None of these
Solution:
(a) Distance between (a cos 25°, 0) and (0, a cos 65°)

Question 3.
If x is a positive integer such that the distance between points P (x, 2) and Q (3, -6) is 10 units, then x =
(a) 3
(b) -3
(c) 9
(d) -9
Solution:
(c) Distance between P (x, 2) and Q (3, -6) = 10 units

=> x (x – 9) + 3 (x – 9) = 0
(x – 9) (x + 3) = 0
Either x – 9 = 0, then x = 9 or x + 3 = 0, then x = -3
x is positive integer
Hence x = 9

Question 4.
The distance between the points (a cosθ + b sinθ, 0) and (0, a sinθ – b cosθ) is
(a) a² + b²
(b) a + b
(c) a² – b²
(d) √(a²+b²)
Solution:
(d) Distance between (a cosθ + b sinθ, 0) and (0, a sinθ – b cosθ)

Question 5.
If the distance between the points (4, p) and (1, 0) is 5, then p =
(a) ±4
(b) 4
(c) -4
(d) 0
Solution:
(a) Distance between (4, p) and (1, 0) = 5

Question 6.
A line segment is of length 10 units. If the coordinates of its one end are (2, -3) and the abscissa of the other end is 10, then its ordinate is
(a) 9, 6
(b) 3, -9
(c) -3, 9
(d) 9, -6
Solution:
(b) Let the ordinate of other end = y
then distance between (2, -3) and (10, y) = 10 units

Question 7.
The perimeter of the triangle formed by the points (0, 0), (1, 0) and (0, 1) is
(a) 1 ± √2
(b) √2 + 1
(c) 3
(d) 2 + √2
Solution:
(d) Let the vertices of ∆ABC be A (0, 0), B(1, 0) and C (0, 1)

Question 8.
If A (2, 2), B (-4, -4) and C (5, -8) are the vertices of a triangle, then the length of the median through vertices C is
(a) √65
(b) √117
(c) √85
(d) √113
Solution:
(c) Let mid point of A (2, 2), B (-4, -4) be D whose coordinates will be

Question 9.
If three points (0, 0), (3, √3) and (3, λ) form an equilateral triangle, then λ =
(a) 2
(b) -3
(c) -4
(d) None of these
Solution:
(d) Let the points (0, 0), (3, √3) and (3, λ) from an equilateral triangle
AB = BC = CA
=> AB² = BC² = CA²
Now, AB² = (x₂ – x₁)² + (y₂ – y₁)²

Question 10.
If the points (k, 2k), (3k, 3k) and (3, 1) are collinear, then k
(a) \(\frac { 1 }{ 3 }\)
(b) \(\frac { -1 }{ 3 }\)
(c) \(\frac { 2 }{ 3 }\)
(d) \(\frac { -2 }{ 3 }\)
Solution:
(b) Let the points A (k, 2k), B (3k, 3k) and C (3, 1) be the vertices of a ∆ABC

Question 11.
The coordinates of the point on x-axis which are equidistant from the points (-3, 4) and (2, 5) are
(a) (20, 0)
(b) (-23, 0)
(c) (\(\frac { 4 }{ 5 }\) , 0)
(d) None of these
Solution:
(d)

Question 12.
If (-1, 2), (2, -1) and (3, 1) are any three vertices of a parallelogram, then
(a) a = 2, b = 0
(b) a = -2, b = 0
(c) a = -2, b = 6
(d) a = 0, b = 4
Solution:
(d) In ||gm ABCD, diagonals AC and AD bisect each other at O
O is mid-point of AC

Question 13.
If A (5, 3), B (11, -5) and P (12, y) are the vertices of a right triangle right angled at P, then y =
(a) -2, 4
(b) -2, 4
(c) 2, -4
(d) 2, 4
Solution:
(c) A (5, 3), B (11, -5) and P (12, y) are the vertices of a right triangle, right angle at P

Question 14.
The area of the triangle formed by (a, b + c), (b, c + a) and (c, a + b) is
(a) a + b + c
(b) abc
(c) (a + b + c)²
(d) 0
Solution:
(d) Vertices of a triangle are (a, b + c), (b, c + a) and (c, a + b)

Question 15.
If (x, 2), (-3, -4) and (7, -5) are coliinear, then x =
(a) 60
(b) 63
(c) -63
(d) -60
Solution:
(c) Area of triangle whose vertices are (x, 2), (-3, -4) and (7, -5)

Question 16.
If points (t, 2t), (-2, 6) and (3, 1) are collinear, then t =
(a) \(\frac { 3 }{ 4 }\)
(b) \(\frac { 4 }{ 3 }\)
(c) \(\frac { 5 }{ 3 }\)
(d) \(\frac { 3 }{ 5 }\)
Solution:
(b) The area of triangle whose vertices are (t, 2t), (-2, 6) and (3, 1)

Question 17.
If the area of the triangle formed by the points (x, 2x), (-2, 6) and (3, 1) is 5 square units, then x =
(a) \(\frac { 2 }{ 3 }\)
(b) \(\frac { 3 }{ 5 }\)
(c) 2
(d) 5
Solution:
(c) Area of triangle whose vertices are (x, 2x), (-2, 6) and (3, 1)

Question 18.
If points (a, 0), (0, b) and (1, 1) are collinear, then \(\frac { 1 }{ a }\) + \(\frac { 1 }{ b }\) =
(a) 1
(b) 2
(c) 0
(d) -1
Solution:
(a) The area of triangle whose vertices are (a, 0), (0, b) and (1, 1)

Question 19.
If the centroid of a triangle is (1, 4) and two of its vertices are (4, -3) and (-9, 7), then the area of the triangle is
(a) 183 sq. units
(b) \(\frac { 183 }{ 2 }\) sq. units
(c) 366 sq. units
(d) \(\frac { 183 }{ 4 }\) sq. units
Solution:
(b) Centroid of a triangle = (1, 4)
and two vertices of the triangle are (4, -3) and (-9, 7)
Let the third vertex be (x, y), then

= \(\frac { 183 }{ 2 }\) sq. units

Question 20.
The line segment joining points (-3, -4) and (1, -2) is divided by y-axis in the ratio
(a) 1 : 3
(b) 2 : 3
(c) 3 : 1
(d) 2 : 3
Solution:
(c) The point lies on y-axis
Its abscissa will be zero
Let the point divides the line segment joining the points (-3, -4) and (1, -2) in the ratio m : n

Question 21.
The ratio in which (4, 5) divides the join of (2, 3) and (7, 8) is
(a) -2 : 3
(b) -3 : 2
(c) 3 : 2
(d) 2 : 3
Solution:
(d) Let the point (4, 5) divides the line segment joining the points (2, 3) and (7, 8) in the ratio m : n

Question 22.
The ratio in which the X-axis divides the segment joining (3, 6) and (12, -3) is
(a) 2 : 1
(b) 1 : 2
(c) -2 : 1
(d) 1 : -2
Solution:
(a) The point lies on x-axis
Its ordinate is zero
Let this point divides the line segment joining the points (3, 6) and (12, -3) in the ratio m : n

Question 23.
If the centroid of the triangle formed by the points (a, b), (b, c) and (c, a) is at the origin, then a³ + b³ + c³ =
(a) abc
(b) 0
(c) a + b + c
(d) 3 abc
Solution:
(d) Centroid of the triangle formed by the points (a, b), (b, c) and (c, a) is origin (0, 0)

Question 24.
If points (1, 2), (-5, 6) and (a, -2) are collinear, then a =
(a) -3
(b) 7
(c) 2
(d) -2
Solution:
(b) The area of a triangle whose vertices are (1, 2), (-5, 6) and (a, -2)

Question 25.
If the centroid of the triangle formed by (7, x), (y, -6) and (9, 10) is at (6, 3), then (x, y) =
(a) (4, 5)
(b) (5, 4)
(c) (-5, -2)
(d) (5, 2)
Solution:
(d) Centroid of (7, x), (y, -6) and (9, 10) is (6, 3)

Question 26.
The distance of the point (4, 7) from the x-axis is
(a) 4
(b) 7
(c) 11
(d) √65
Solution:
(b) The distance of the point (4, 7) from x-axis = 7

Question 27.
The distance of the point (4, 7) from the y-axis is
(a) 4
(b) 7
(c) 11
(d) √65
Solution:
(a) The distance of the point (4, 7) from y-axis = 4

Question 28.
If P is a point on x-axis such that its distance from the origin is 3 units, then the coordinates of a point Q on OY such that OP = OQ, are
(a) (0, 3)
(b) (3, 0)
(c) (0, 0)
(d) (0, -3)
Solution:
(a) P is a point on x-axis and its distance from 0 is 3
Co-ordinates of P will be (3, 0)
Q is a point on OY such that OP = OQ
Co-ordinates of Q will be (0, 3)

Question 29.
If the point (x, 4) lies on a circle whose centre is at the origin and radius is 5, then x =
(a) ±5
(b) ±3
(c) 0
(d) ±4
Solution:
(b) Point A (x, 4) is on a circle with centre O (0, 0) and radius = 5

Question 30.
If the point P (x, y) is equidistant from A (5, 1) and B (-1, 5), then
(a) 5x = y
(b) x = 5y
(c) 3x = 2y
(d) 2x = 3y
Solution:
(c)

Question 31.
If points A (5, p), B (1, 5), C (2, 1) and D (6, 2) form a square ABCD, then p =
(a) 7
(b) 3
(c) 6
(d) 8
Solution:
(c) Vertices of a square are A (5, p), B (1, 5), C (2, 1) and D (6, 2)
The diagonals bisect each other at O
O is the mid-point of AC and BD
O is mid-point of BD, then

Question 32.
The coordinates of the circumcentre of the triangle formed by the points O (0, 0), A (a, 0) and B (0, b) are
(a) (a, b)
(b) (\(\frac { a }{ 2 }\) , \(\frac { b }{ 2 }\))
(c) (\(\frac { b }{ 2 }\) , \(\frac { a }{ 2 }\))
(d) (b, a)
Solution:
(b) Let co-ordinates of C be (x, y) which is the centre of the circumcircle of ∆OAB
Radii of a circle are equal

Question 33.
The coordinates of a point on x-axis which lies on the perpendicular bisector of the line segment joining the points (7, 6) and (-3, 4) are
(a) (0, 2)
(b) (3, 0)
(b) (0, 3)
(d) (2, 0)
Solution:
(d) The given point P lies on x-axis
Let the co-ordinates of P be (x, 0)
The point P lies on the perpendicular bisector of of the line segment joining the points A (7, 6), B (-3, 4)

Question 34.
If the centroid of the triangle formed by the points (3, -5), (-7, 4), (10, -k) is at the point (k, -1), then k =
(a) 3
(b) 1
(c) 2
(d) 4
Solution:
(c) O (k, -1) is the centroid of triangle whose vertices are

Question 35.
If (-2, 1) is the centroid of the triangle having its vertices at (x, 0), (5, -2), (-8, y), then x, y satisfy the relation
(a) 3x + 8y = 0
(b) 3x – 8y = 0
(c) 8x + 3y = 0
(d) 8x = 3y
Solution:
(-2, 1) is the centroid of triangle whose vertices are (x, 0), (5, -2), (-8, y)

Question 36.
The coordinates of the fourth vertex of the rectangle formed by the points (0, 0), (2, 0), (0, 3) are
(a) (3, 0)
(b) (0, 2)
(c) (-2, 3)
(d) (3, 2)
Solution:
(c) Three vertices of a rectangle are A (0, 0), B (2, 0), C (0, 3)
Let fourth vertex be D (x, y)
The diagonals of a rectangle bisect eachother at O
O is the mid-point of AC, then
Coordinates of O will be (\(\frac { 0+0 }{ 2 }\) , \(\frac { 0+3 }{ 2 }\))

Question 37.
The length of a line segment joining A (2, -3) and B is 10 units. If the abscissa of B is 10 units, then its ordinates can be
(a) 3 or -9
(b) -3 or 9
(c) 6 or 27
(d) -6 or-27
Solution:
(a) Abscissa of B is 10 and co-ordinates of A are (2, -3)
Let ordinates of B be y, then

Question 38.
The ratio in which the line segment joining P(x₁, y₁) and Q (x₂, y₂) is divided by x-axis is
(a) y₁ : y₂
(b) -y₁ : y₂
(c) x₁ : x₂
(d) -x₁ : x₂
Solution:
(b) Let a point A on x-axis divides the line segment joining the points P (x₁, y₁), Q (x₂, y₂) in the ratio m₁ : m₂ and
let co-ordinates of A be (x, 0)

Question 39.
The ratio in which the line segment joining points A (a₁, b₁) and B (a₂, b₂) is divided by y-axis is
(a) -a₁ : a₂
(b) a₁ : a₂
(c) b₁ : b₂
(d) -b₁ : b₂
Solution:
(a) Let the point P on y-axis, divides the line segment joining the point A (a₁, b₁) and B (a₂, b₂) is the ratio m₁ : m₂ and
let the co-ordinates of P be (0, y), then

Question 40.
If the line segment joining the points (3, -4) and (1, 2) is trisected at points P

Solution:
(b)

Question 41.
If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (-2, 5), then the coordinates of the other end of the diameter are [CBSE 2012]
(a) (-6, 7)
(b) (6, -7)
(c) (6, 7)
(d) (-6, -7)
Solution:
(a) Let AB be the diameter of a circle with centre O

Question 42.
The coordinates of the point P dividing the line segment joining the points A (1, 3) and B (4, 6) in the ratio 2 : 1 are
(a) (2, 4)
(b) (3, 5)
(c) (4, 2)
(d) (5, 3) [CBSE 2012]
Solution:
(b) Point P divides the line segment joining the points A (1, 3) and B (4, 6) in the ratio 2 : 1
Let coordinates of P be (x, y), then

Question 43.
In the figure, the area of ∆ABC (in square units) is [CBSE 2013]
(a) 15
(b) 10
(c) 7.5
(d) 2.5

Solution:
(c)

Question 44.
The point on the x-axis which is equidistant from points (-1, 0) and (5, 0) is
(a) (0, 2)
(b) (2, 0)
(c) (3, 0)
(d) (0, 3) [CBSE 2013]
Solution:
(c) Let the point P (x, 0) is equidistant from the points A (-1, 0), B (5, 0)

Question 45.
If A (4, 9), B (2, 3) and C (6, 5) are the vertices of ∆ABC, then the length of median through C is
(a) 5 untis
(b) √10 units
(c) 25 units
(d) 10 units [CBSE 2014]
Solution:
(b) A (4, 9), B (2, 3) and C (6, 5) are the vertices of ∆ABC
Let median CD has been drawn C (6, 5)

Question 46.
If P (2, 4), Q (0, 3), R (3, 6) and S (5, y) are the vertices of a paralelogram PQRS, then the value of y is
(a) 7
(b) 5
(c) -7
(d) -8 [CBSE 2014]
Solution:
(a) P (2, 4), Q (0, 3), R (3, 6) and S (5, y) are the vertices of a ||gm PQRS

Question 47.
If A (x, 2), B (-3, -4) and C (7, -5) are collinear, then the value of x is
(a) -63
(b) 63
(c) 60
(d) -60 [CBSE 2014]
Solution:
(a) A (x, 2), B (-3, -4) and C (7, -5) are collinear, then area ∆ABC = 0
Now area of ∆ABC

Question 48.
The perimeter of a triangle with vertices (0, 4) and (0, 0) and (3, 0) is
(a) 7 + √5
(b) 5
(c) 10
(d) 12 [CBSE 2014]
Solution:
(d) A (0, 4) and B (0, 0) and C (3, 0) are the vertices of ∆ABC

Question 49.
If the point P (2, 1) lies on the line segment joining points A (4, 2) and B (8, 4), then
(a) AP = \(\frac { 1 }{ 3 }\) AB
(b) AP = BP
(c) BP = \(\frac { 1 }{ 3 }\) AB
(d) AP = \(\frac { 1 }{ 2 }\) AB
Solution:
(d) Given that, the point P (2, 1) lies on the line segment joining the points (4, 2) and B (8, 4), which shows in the figure below:

Question 50.
A line intersects the y-axis and x-axis at P and Q, respectively. If (2, -5) is the mid-point of PQ, then the coordinates of P and Q are, respectively
(a) (0, -5) and (2, 0)
(b) (0, 10) and (-4, 0)
(c) (0, 4) and (-10, 0)
(d) (0, -10) and (4, 0)
Solution:
(d) Let the coordinates of P and Q (0, y) and (x, 0), respectively.
So, the mid-point of P (0, y) and Q (x, 0) is

2 = \(\frac { x + 0 }{ 2 }\) and -5 = \(\frac { y + 0 }{ 2 }\)
=> 4 = x and -10 = y
=> x = 4 and y = -10
So, the coordinates of P and Q are (0, -10) and (4, 0).

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.1
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

Answer each of the following questions either in one word or one sentence or as per requirement of the questions :
Question 1.
Write the distance between the points A (10 cosθ, 0) and B (0, 10 sinθ).
Solution:
Distance between the points A (10 cosθ, 0) and B (0, 10 sinθ)

Question 2.
Write the perimeter of the triangle formed by the points O (0, 0), A (a, 0), and B (0, b).
Solution:
The vertices of a ∆OAB, O (0, 0), A (a, 0), and B (0, b)
Now length of OA

Question 3.
Write the ratio in which the line segment joining points (2, 3) and (3, -2) is divided by x-axis.
Solution:
The required point is on x-axis
Its ordinate will be 0
Let the point be (x, 0) and let this point divides the join of the points (2, 3) and (3, -2) in the ratio m : n

Question 4.
What is the distance between the points (5 sin 60°, 0) and (0, 5 sin 30°) ?
Solution:
Distance between the given points

Question 5.
If A (-1, 3), B (1, -1) and C (5, 1) are the vertices of a triangle ABC, what is the length of the median through vertex A ?
Solution:
The vertices of ∆ABC are A (-1, 3), B (1, -1) and C (5, 1)
Let AD be the median

Question 6.
If the distance between points (x, 0) and (0, 3) is 5, what are the value of x ?
Solution:
Distance between (x, 0) and (0, 3) = 5

Question 7.
What is the area of the triangle formed by the points O (0, 0), A (6, 0) and B (0, 4) ?
Solution:
The vertices of the triangle OAB are O (0, 0), A (6, 0) and B (0, 4)

Question 8.
Write the coordinates of the point dividing line segment joining points (2, 3) and (3, 4) internally in the ratio 1 : 5.
Solution:
Let the coordinates of the required point be (x, y), then

Question 9.
If the centroid of the triangle formed by points P (a, b), Q (b, c) and R (c, a) is at the origin, what is the value of a + b + c ?
Solution:
Vertices of ∆PQR are P (a, b), Q (b, c) and R (c, a) and its centroid = O (0, 0)

Question 10.

Solution:

Question 11.
Write the coordinates of a point on x- axis which is equidistant from the points (-3, 4) and (2, 5).
Solution:
The point is on x-axis
Its ordinates of the point P is (x, 0)
P is equidistant from A (-3, 4) and B (2, 5)

Question 12.
If the mid-point of the segment joining A (x, y + 1) and B (x + 1, y + 2) is C (\(\frac { 3 }{ 2 }\) , \(\frac { 5 }{ 2 }\)) find x, y.
Solution:
C (\(\frac { 3 }{ 2 }\) , \(\frac { 5 }{ 2 }\)) is mid point of the line segment

Question 13.
Two vertices of a triangle have co-ordinates (-8, 7) and (9, 4). If the centroid of the triangle is at the origin, what are the co-ordinates of the third vertex ?
Solution:
Two vertices of a triangle are (-8, 7) and (9, 4)
Let the third vertex be (x, y)
Centroid of the triangle is (0, 0)

Question 14.
Write the coordinates the reflections of points (3, 5) in x and y-axes.
Solution:
Reflection of P (3, 5) in x-axis is will be (3, -5)
and reflection of P in y-axis will be (-3, 5)

Question 15.
If points Q and R reflections of point P (-3, 4) in X and Y axes respectively, what is QR ?
Solution:
Reflection of point P (-3, 4) in X-axis will be Q with coordinates Q (-3, -4) and reflection in Y-axis will be R (3, 4)

Question 16.
Write the formula for the area of the triangle having its vertices at (x₁, y₁), (x₂, y₂) and (x₃, y₃).
Solution:

Question 17.
Write the condition of collinearity of points (x₁, y₁), (x₂, y₂) and (x₃, y₃).
Solution:
Three points (x₁, y₁), (x₂, y₂) and (x₃, y₃). are said to be collinear if the area of the triangle formed by these point = 0 i.e.,

Question 18.
Find the values of x for which the distance between the point P (2, -3) and Q (x, 5) is 10.
Solution:
Distance between P (2, -3) and Q (x, 5) = 10

Question 19.
Write the ratio in which the line segment joining the points A (3, -6) and B (5, 3) is divided by X-axis.
Solution:
The point lies on x-axis
Its ordinate will be = 0
Let the point P (x, 0) divides the line segment joining the points A (3, -6) and B (5, 3) in the ratio m : n.

Question 20.
Find the distance between the points (\(\frac { -8 }{ 5 }\) , 2) and (\(\frac { 2 }{ 5 }\) , 2). (C.B.S.E. 2009)
Solution:

Question 21.
Find the value of a so that the point (3, a) lies on the line represented by 2x – 3y + 5 = 0. (C.B.S.E. 2009)
Solution:

Question 22.
What is the distance between the points A (c, 0) and B (0, – c) ? [CBSE 2010]
Solution:

Question 23.
If P (2, 6) is the mid-point of the line segment joining A (6, 5) and B (4, y), find y. [CBSE 2010]
Solution:
P (2, 6) is the mid-point of the line segment A (6, 5) and b (4, y)

Question 24.
If the distance between the points (3, 0) and (0, y) is 5 units and y is positive, then what is the value of y ? [CBSE 2010]
Solution:
Distance between (3, 0) and (0, y) is 5 units

Question 25.
If P (x, 6) is the mid-point of the line segment joining A (6, 5) and B (4, y), find y. [CBSE 2010]
Solution:
P (x, 6) is the mid-point of the line segment joining the points A (6, 5), B (4, y)

Question 26.
If P (2, p) is the mid-point of the line segment joining the points A (6, -5) and B (-2, 11), find the value of p. [CBSE 2010]
Solution:
P (2, p) is the mid-point of the line segment joining the points A (6, -5) and B (-2, 11)

Question 27.
If A (1, 2), B (4, 3) and C (6, 6) are the three vertices of a parallelogram ABCD, find the coordinates of fourth vertex D. [CBSE 2010]
Solution:
vertices of a parallelogram Let co-ordinates of D be (x, y)
Diagonals AC and BD bisect each other at O
Co-ordinates of O will be

Question 28.
What is the distance between the points A (sinθ – cosθ, 0) and B (0, sinθ + cosθ)? [CBSE 2015]
Solution:

Question 29.
What are the coordinates of the point where the perpendicular bisector of the line segment joining the points A (1, 5) and B (4, 6) cuts the y-axis?
Solution:
Firstly, we plot the points of the line segment on the paper and join them.


Now, we draw a straight line on paper passes through the mid-point P.
We see that the perpendicular bisector cuts the y-axis at the point (0, 13).
Hence, the required point is (0, 13).

Question 30.
Find the area of the triangle with vertices (a, b + c), (b, c + a) and (c, a + b).
Solution:

Question 31.
If the points A (1, 2), O (0, 0) and C (a, b) are collinear, then find a : b.
Solution:

=> 2a = b
Hence, the required relation is 2a = b

Question 32.
Find the coordinates of the point which is equidistant from the three vertices A (2x, 0), O (0, 0) and B (0, 2y) of ∆AOB.
Solution:
Let the coordinate of the point which is equidistant from the three vertices O (0, 0), A (0, 2y) and B (2x, 0) is P (h, k).

Question 33.
If the distance between the points (4, k), and (1, 0) is 5, then what can be the possible value of k? [CBSE 2017]
Solution:
Let the points x (4, k) and y (1, 0)
It is given that the distance xy is 5 units.
By using the distance formula,

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.1
• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.2
• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.3
• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles Ex 13.4
• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS
• RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles MCQS

Answer each of the following questions either in one word or one sentence or as per requirement of the questions :
Question 1.
What is the ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal?
Solution:
Diameter of a circle and side of an equilateral triangle are same
Let the diameter of the circle = a
Then radius (r) = \(\frac { a }{ 2 }\)

Question 2.
If the circumference of two circles are in the ratio 2 : 3, what is the ratio of their areas ?
Solution:
Let R and r be the radii of two circles, then the ratio between their circumferences = 2πR : 2πr

Question 3.
Write the area of the sector of a circle whose radius is r and length of the arc is l.
Solution:
Let arc l subtends angle 9 at the centre of the circle
Now radius of a circle = r
and length of arc =l

Question 4.
What is the length (in terms of π) of the arc that subtends an angle of 36° at the centre of a circle of radius 5 cm?
Solution:
Radius of the circle = 5 cm
Angle at the center = 36°

Question 5.
What is the angle subtended at the centre of a circle of radius 6 cm by an arc of length 3π cm ?
Solution:
Let the arc subtends angle θ at the centre of a circle
Radius of circle (r) = 6 cm
Length of arc = 3π cm

Question 6.
What is the area of a sector of a circle of radius 5 cm formed by an arc of length 3.5 cm ?
Solution:
Radius of the circle (r) = 5 cm
Length of arc (l) = 3.5 cm
Let angle 9 be subtended by the arc at the centre

Question 7.
In a circle of radius 10 cm, an arc subtends an angle of 108° at the centre. What is the area of the sector in terms of π ?
Solution:
Radius of the circle = 10 cm
Angle at the centre = 108°

Question 8.
If a square is inscribed in a circle, what is the ratio of the areas of the circle and the square ?
Solution:
A square ABCD is inscribed in a circle with centre O

Let the radius of the circle = r
Then its area = πr²
Now diagonal of the square = diameter of the circle = 2r

Question 9.
Write the formula for the area of a sector of angle θ (in degrees) of a circle of radius r.
Solution:
Area of a sector of a circle whose radius = r

Question 10.
 Write the formula for the area of a segment in a circle of radius r given that the sector angle is 0 (in degrees).
Solution:
Radius of the circle = r
and angle subtended by the sector at the centre = θ
Area of the segment

Question 11.
If the adjoining figure is a sector of a circle of radius 10.5 cm, what is the perimeter of the sector ? (Take π= 22/7)

Solution:
Radius of the circle = 10.5 cm
Angle at the centre of the circle = 60°

Question 12.
If the diameter of a semi-circular protractor is 14 cm then find its perimeter. (C.B.S.E. 2009)
Solution:
Diameter of semicircular protractor = 14 cm

∴ Radius (r) =  \(\frac { 14 }{ 2 }\) = 7 cm
Now perimeter of protractor

Question 13.
An arc subtends an angle of 90° at the centre of the circle of radius 14 cm. Write the area of minor sector thus formed in terms of π.
Solution:
AB is an arc of the circle with centre O and radius 14 cm and subtends an angle of 90° at the centre O.

Question 14.
Find the area of the largest triangle that can be inscribed in a semi-circle of radius r units. [CBSE 2015]
Solution:
Radius of semicircle = r

In semicircle ΔABC is the largest triangle whose base is AC = 2 x r = 2r units
and height OB = r units

Question 15.
Find the area of a sector of circle of radius 21 cm and central angle 120°.
Solution:

Question 16.
What is the area of a square inscribed in a circle of diameter p cm?
Solution:
Diameter AC of the circle is p.
Also AC is diagonal of square ABCD.
Each angle of square is of 90°

Question 17.
Is it true to say that area of a segment of a circle is less than the area of its corresponding sector? Why?
Solution:
False.
It is true only in the case of minor segment. But in case of major segment, area is always greater than the area of sector.

Question 18.
If the numerical value of the area of a circle is equal to the numerical value of its circumference, find its radius.
Solution:
∵ Numerical value of area of circle = Numerical value of circumference
∴  πr² = 2πr
or r = 2 units

Question 19.
How many revolutions a circular wheel of radius r metres makes in covering a distance of s metres?
Solution:
Radius of circular of wheel (r) = r m
Circumference of a circular wheel = 2πr
Distance to be covered = Sm

Question 20.
Find the ratio of the area of the circle circumscribing a square to the area of the circle inscribed in the square.
Solution:
Let each side of of square = x
∴ Diameter of inner circle = x
Radius r = \(\frac { x }{ 2 }\)
Diameter of outer circle = AD

Hope given RD Sharma Class 10 Solutions Chapter 13 Areas Related to Circles VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5

Other Exercises

• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.1
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

Question 1.
Find the area of a triangle whose vertices are :
(i) (6, 3), (-3, 5) and (4, -2)
(ii) (\({ at }_{ 1 }^{ 2 }\), 2at₁), (\({ at }_{ 2 }^{ 2 }\), 2at₂) and (\({ at }_{ 3 }^{ 2 }\), 2at₃)
(iii) (a, c + a), (a, c) and (-a, c – a)
Solution:
(i) Co-ordinates of ∆ABC are A (6, 3), B (-3, 5) and C (4, -2)

Question 2.
Find the area of the quadrilaterals, the coordinates of whose vertices are
(i) (-3, 2), (5, 4), (7, -6) and (-5, -4)
(ii) (1, 2), (6, 2), (5, 3) and (3, 4)
(iii) (-4, -2), (-3, -5), (3, -2), (2, 3) (C.B.S.E. 2009)
Solution:
(i) Let vertices of quadrilateral ABCD be A (-3, 2), B (5, 4), C (7, -6) and D (-5, -4)
Join AC

Question 3.
The four vertices of a quadrilaterals are (1, 2), (-5, 6), (7, -4) and (k, -2) taken in order. If the area of the quadrilateral is zero, find the value of k ?
Solution:
Let the vertices of quadrilateral ABCD be
A (1, 2), B (-5, 6), C (7, -4) and D (k, -2)
Join AC

Question 4.
The vertices of ∆ABC are (-2, 1), (5, 4) and (2, -3) respectively. Find the area of the triangle and the length of the altitude through A.
Solution:
Vertices of ∆ABC are A (-2, 1), B (5, 4) and C (2, -3) and AD ⊥ BC, let AD = h
Now area of ∆ABC

Question 5.
Show that the following sets of points are collinear
(a) (2, 5), (4, 6) and (8, 8)
(b) (1, -1), (2, 1) and (4, 5)
Solution:
We know that points are collinear if the area of the triangle formed by them is zero
(a) Vertices of ∆ABC are (2, 5), (4, 6) and (8, 8)

Question 6.
Find the area of a quadrilateral ABCD, the coordinates of whose varities are A (-3, 2), B (5, 4), C (7, -6) and D (-5, -4). [CBSE 2016]
Solution:
Area of quadrilateral ABCD
= area of ∆ABC + area of ∆ACD

Question 7.
In ∆ABC, the coordinates of vertex A are (0, -1) and D (1, 0) and E (0, 1) respectively the mid-points of the sides AB and AC. If F is the mid-point of side C, find the area of ∆DEF. [CBSE 2016]
Solution:
Let B (p, q), C (r, s) and F (x, y)
Mid-point of AB = Coordinates of D

Question 8.
Find the area of the triangle PQR with Q (3, 2) and the mid-points of the sides through Q being (2, -1) and (1, 2). [CBSE 2015]
Solution:
In ∆PQR, L and N are mid points of QR and QP respectively coordinates of Q are (3, 2) of L are (2, -1) and of N are (1, 2)

Question 9.
If P (-5, -3), Q (-4, -6), R (2, -3) and S (1, 2) are the vertices of a quadrilateral PQRS, find its area. [CBSE 2015]
Solution:
P (-5, -3), Q (-4, -6), R (2, -3) and S (1,2) are the vertices of a quadrilateral PQRS
Join PR which forms two triangles PQR and PSR

Question 10.
If A (-3, 5), B (-2, -7), C (1, -8) and D (6, 3) are the vertices of a quadrilateral ABCD, find its area. [CBSE 2014]
Solution:
A (-3, 5), B (-2, -7), C (1,-8) and D (6, 3) are the vertices of a quadrilateral ABCD
Join AC

Question 11.
For what value of ‘a’ the points (a, 1), (1, -1) and (11, 4) are collinear?
Solution:
Let the vertices of ∆ABC are A (a, 1), B (1, -1) and C (11, 4)

Question 12.
Prove that the points (a, b), (a₁, b₁) and (a – a₁, b – b₁) are collinear if ab₁ = a₁b.
Solution:

Question 13.
If the vertices of a triangle are (1, -3), (4, p) and (-9, 7) and its area is 15 sq. units, find the value(s) of p. [CBSE 2012]
Solution:
The vertices of a triangle are (1, -3), (4, p) and (-9, 7) and area of triangle = 15 sq. units

Question 14.
If (x, y) be on the line joining the two points (1, -3) and (-4, 2), prove that x + y + 2 = 0.
Solution:
Point (x, y) be on the line joining the two points (1, -3) and (-4, 2)
Points (x, y), (1, -3) and (-4, 2) are collinear
Let the points (x, y) (1, -3) and (-4, 2) are the vertices of a triangle, then

Question 15.
Find the value of k if points (k, 3), (6, -2) and (-3, 4) are collinear. [CBSE 2008]
Solution:
Let the points (k, 3), (6, -2) and (-3, 4) be the vertices of a triangle, then

Question 16.
Find the value of k, if the points A (7, -2), B (5, 1) and C (3, 2k) are collinear. [CBSE 2010]
Solution:
Points A (7, -2), B (5, 1) and C (3, 2k) are collinear
area of ∆ABC = 0

Question 17.
If the point P (m, 3) lies on the line segment joining the points A (\(\frac { -2 }{ 5 }\) , 6) and B (2, 8), find the value of m.
Solution:

Question 18.
If R (x, y) is a point on the line segment joining the points P (a, b) and Q (b, a), then prove that x + y = a + b. [CBSE 2010]
Solution:
Point R (x, y) lies on the line segment joining the points P (a, b) and Q (b, a)
Area of ∆PRQ = 0

Question 19.
Find the value of k, if the points A (8, 1), B (3, -4) and C (2, k) are collinear. [CBSE 2010]
Solution:
The points A (8, 1), B (3, -4) and C (2, k) are collinear
Area of ∆ABC = 0

Question 20.
Find the value of a for which the area of the triangle formed by the points A (a, 2a), B (-2, 6) and C (3, 1) is 10 square units.
Solution:

Question 21.
If a ≠ b ≠ 0, prove that the points (a, a²), (b, b²), (0, 0) are never collinear. [CBSE 2017]
Solution:

Question 22.
The area of a triangle is 5 sq. units. Two of its vertices are at (2, 1) and (3, -2). If the third vertex is (\(\frac { 7 }{ 2 }\) , y), find y. [CBSE 2017]
Solution:

Question 23.
Prove that the points (a, 0), (0, b) and (1, 1) are collinear if, \(\frac { 1 }{ a }\) + \(\frac { 1 }{ b }\) = 1.
Solution:
Let the points are A (a, 0), B (0, b) and C (1, 1) which form a triangle

Question 24.
The point A divides the join of P (-5, 1) and Q (3, 5) in the ratio k : 1. Find the two values of k for which the area of ∆ABC where B is (1, 5) and C (7, -2) is equal to 2 units.
Solution:
Let the coordinates of A be (x, y) which divides the join of P (-5, 1) and Q (3, 5) in the ratio. Then

Question 25.
The area of a triangle is 5. Two of its vertices are (2, 1) and (3, -2). The third vertex lies on y = x + 3. Find the third vertex.
Solution:
Let the coordinates of third vertex of the triangle be (x, y) and other two vertices are (2, 1) and (3, 2)

Question 26.
If a ≠ b ≠ c, prove that the points (a, a²), (b, b²), (c, c²) can never be collinear.
Solution:

Question 27.
Four points A (6, 3), B (-3, 5), C (4, -2) and D (x, 3x) are given in such a way that \(\frac { \triangle DBC }{ \triangle ABC } =\frac { 1 }{ 2 }\) , find x?
Solution:
Let A (6, 3), B (-3, 5), C (4, -2) and D (x, 3x) are the vertices of quadrilateral ABCD
AC and BD are joined

Question 28.
If three points (x₁, y₁), (x₂, y₂), (x₃, y₃) lie on the same line, prove that

Solution:
Let the points (x₁, y₁), (x₂, y₂), (x₃, y₃) are the vertices of a triangle

Question 29.
Find the area of a parallelogram ABCD if three of its vertices are A (2, 4), B (2 + √3, 5) and C (2, 6). [CBSE 2013]
Solution:
Three vertices of a ||gm ABCD are A (2, 4), B (2 + √3 , 5) and C (2, 6).
Draw one diagonal AC of ||gm ABCD

Question 30.
Find the value (s) of k for which the points (3k – 1, k – 2), (k, k – 7) and (k – 1, -k – 2) are collinear. [CBSE 2014]
Solution:

Question 31.
If the points A (-1, -4), B (b, c) and C (5, -1) are collinear and 2b + c = 4, find the values of b and c. [CBSE 2014]
Solution:

Question 32.
If the points A (-2, 1), B (a, b) and C (4, -1) are collinear and a – b = 1, find the values of a and 6. [CBSE 2014]
Solution:
Points A (-2, 1), B (a, b) and C (4, -1) are
collinear if area ∆ABC = 0
Now area of ∆ABC

Question 33.
If the points A (1, -2), B (2, 3), C (a, 2) and D (-4, -3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base. [NCERT Exemplar]
Solution:
In parallelogram, we know that, diagonals bisects each other
i.e., mid-point of AC = mid-point of BD

Question 34.
A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD. If E is the mid-point of DC, find the area of ∆ADE. [NCERT Exemplar]
Solution:
Given that, A (6,1), B (8,2) and C (9,4) are three vertices of a parallelogram ABCD.
Let the fourth vertex of parallelogram be (x, y).
We know that, the diagonal of a parallelogram bisect each other.

Question 35.
If D (\(\frac { -1 }{ 2 }\), \(\frac { 5 }{ 2 }\)) E (7, 3) and F (\(\frac { 7 }{ 2 }\), \(\frac { 7 }{ 2 }\)) are the mid-points of sides of ∆ABC, find the area of ∆ABC. [NCERT Exemplar]
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4

Other Exercises

• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.1
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS
• RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

Question 1.
Find the centroid of the triangle whose vertices are :
(i) (1, 4), (-1, -1), (3, -2)
(ii) (-2, 3), (2, -1), (4, 0)
Solution:

Question 2.
Two vertices of a triangle are (1, 2), (3, 5) and its centroid is at the origin. Find the Co-ordinates of the third vertex.
Solution:
Centroid of a triangle is O(0, 0) ….(i)
Co-ordinates of two vertices of a ∆ABC are A (1, 2) and B (3, 5)
Let the third vertex be (x, y)

Question 3.
Find the third vertex of a triangle, if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin.
Solution:
Let two vertices of a ∆ABC be A (-3, 1) and B (0, -2) and third vertex C be (x, y)
Centroid of the ∆ABC is O (0, 0)

Question 4.
A (3, 2) and B (-2, 1) are two vertices of a triangle ABC whose centroid G has the coordinates (\(\frac { 5 }{ 3 }\) , \(\frac { -1 }{ 3 }\)) . Find the coordinates of the third vertex C of the triangle. [CBSE 2004]
Solution:
A (3, 2) and B (-2, 1) are the two vertices of ∆ABC whose centroid is G (\(\frac { 5 }{ 3 }\) , \(\frac { -1 }{ 3 }\))
Let third vertex C be (x, y)

Question 5.
If (-2, 3), (4, -3) and (4, 5) are the mid-points of the sides of a triangle, find the co-ordinates of its centroid.
Solution:
In ∆ABC, D, E and F are the mid-points of the sides BC, CA and AB respectively.
The co-ordinates of D are (-2, 3), of E are (4,-3) and of F are (4, 5)
Let the co-ordinates of A, B and C be (x₁, y₁), (x₂, y₂), (x₃, y₃) respectively

Question 6.
Prove analytically that the line segment joining the middle points of two sides of a triangle is equal to half of the third side.
Solution:
In ∆ABC,
D and E are the mid points of the sides AB and AC respectively

Question 7.
Prove that the lines joining the middle points of the opposite sides of a quadrilateral and the join of the middle points of its diagonals meet in a point and bisect one another.
Solution:
Let A (x₁, y₁), B (x₂, y₂), C (x₃, y₃) and D (x₄, y₄) be the vertices of quadrilateral ABCD
E and F are the mid points of side BC and AD respectively and EF is joined G and H are the mid points of diagonal AC and BD.
GH are joined

Question 8.
If G be the centroid of a triangle ABC and P be any other point in the plane, prove that PA² + PB² + PC² = GA² + GB² + GC² + 3GP².
Solution:
In AABC, G is the centroid of it Let P (h, x) is any point in the plane
Let co-ordinates of A are (x₁, y₁) of B are (x₂, y₂) and of C are (x₃, y₃)


Hence proved.

Question 9.
If G be the centroid of a triangle ABC, prove that AB² + BC² + CA² = 3 (GA² + GB² + GC²)
Solution:
Let the co-ordinates of the vertices of ∆ABC be A (x₁, y₁), B (x₂, y₂), C (x₃, y₃) and let G be the centroid of the triangle



Hence proved.

Question 10.
In the figure, a right triangle BOA is given. C is the mid-point of the hypotenuse AB. Show that it is equidistant from the vertices O, A and B.
Solution:

In right ∆OAB, co-ordinates of O are (0, 0) of A are (2a, 0) and of B are (0, 2b)
C is the mid-point of AB
Co-ordinates of C will be

We see that CO = CA = CB
Hence C is equidistant from the vertices O, A and B.
Hence proved.

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.