RS Aggarwal Maths Solutions Class 10

RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Choose the correct answer in each of the following questions.
Question 1.
Solution:
Distance between P (-6, 8) and origin O (0, 0)

Question 2.
Solution:
Distance of the point (-3, 4) from x-axis = 4 (c)

Question 3.
Solution:
Let point P (x, 0) is on x-axis which is equidistant from points A (-1, 0) and B (5, 0)

Question 4.
Solution:
R (5, 6) is the midpoint of the line segment joining the points A (6, 5) and B (4, y)

Question 5.
Solution:
Point C (k, 4) divides the join of points A (2, 6) and B (5, 1) in the ratio 2 : 3

Question 6.
Solution:
Vertices of ∆ABC are A (0, 4), B (0, 0) and C (3, 0)

Question 7.
Solution:
A (1, 3), B (-1, 2), C (2, 5) and D (x, 4) are the vertices of a ||gm ABCD.
AB = CD

Question 8.
Solution:
Points A (x, 2), B (-3, -4) and C (7, -5) are collinear

Question 9.
Solution:
Area of ∆ABC whose vertices are A (5, 0), B (8, 0) and C (8, 4)

Question 10.
Solution:
Area of ∆ABC with vertices A (a, 0), O (0, 0) and B (0, b)

Question 11.
Solution:
P( \(\frac { a }{ 2 }\), 4) is midpoint of the line segment joining the points A (-6, 5) and B (-2, 3)

Question 12.
Solution:
ABCD is a rectangle whose three vertices are B (4, 0), C (4, 3) and D (0, 3)

Question 13.
Solution:
Let coordinates of P be (x, 7)
P divides the line segment joining the points A (1, 3) and B (4, 6) in the ratio 2 : 1

Question 14.
Solution:
Let coordinates of one end of diameter of a circle are A (2, 3) coordinates of centre are (-2, 5)
Let coordinates of other end B of the diameter be (x, y)

Question 15.
Solution:
In the given figure, P (5, -3) and Q (3, y) are the points of trisection of the line segment joining A (7, -2) and B (1, -5)

Question 16.
Solution:
Midpoint of AB is P (0, 4) coordinates of B are (-2, 3)

Question 17.
Solution:
Let point P (x, y) divides AB with vertices A(2, -5) and B (5, 2) in the ratio 2 : 3

Question 18.
Solution:
A (-6, 7) and B (-1, -5) two points, then

Question 19.
Solution:
Let point P (x, 0) on x-axis is equidistant from A (7, 6) and B (-3, 4)

Question 20.
Solution:
Distance of P (3, 4) from the x-axis = 4 units. (b)

Question 21.
Solution:
Let a point P (x, 0) on x-axis divides the line segment.
AB joining the points A (2, -3) and B (5, 6) in the ratio m : n.

Question 22.
Solution:
Let a point A (0, y) on y-axis divides the line segment PQ joining the points P (-4, 2) and Q (8, 3) in the ratio of m : n.

Question 23.
Solution:
P (-1, 1) is the midpoint of line segment joining the points A (-3, b) and B (1, b + 4)

Question 24.
Solution:
Let the point P (x, y) divides the line segment joining the points A (2, -2) and B (3, 7) in the ratio k : 1, then

Question 25.
Solution:
AD is the median of ∆ABC with vertices A (4, 2), B (6, 5) and C (1, 4)

Question 26.
Solution:
A (-1, 0), B (5, -2) and C (8, 2) are the vertices of a ∆ABC then centroid

Question 27.
Solution:
Two vertices A (-1, 4), B (5, 2) of a ∆ABC and its centroid G is (0, -3)
Let (x, y) be the co-ordinates of vertex C, then

Question 28.
Solution:
Points A (-4, 0), B (4, 0) and C (0, 3) are vertices of a ∆ABC.

Question 29.
Solution:
Points P (0, 6), Q (-5, 3) and R (3, 1) are the vertices of a ∆PQR

Question 30.
Solution:
Points A (2, 3), B (5, k) and C (6, 7) are collinear
Area of ∆ABC = 0

Question 31.
Solution:
Points A (1, 2), B (0, 0) and C (a, b) are collinear
Area of ∆ABC = 0

Question 32.
Solution:
A (3, 0), B (7, 0) and C (8, 4)
Area ∆ABC

Question 33.
Solution:
AOBC is a rectangle with vertices A (0, 3), O (0, 0) and B (5, 0) and each diagonal of rectangle are equal.

Question 34.
Solution:
Points are A (4, p) and B (1, 0)
Distance = 5 units

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS 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.

RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry Ex 16D

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Very-Short-Answer Questions
Question 1.
Solution:
Points are A (-1, y), B (5, 7) and centre O (2, -3y).
Points A and B lie on the circle with centre O.

or y + 1 = 0, then y = -1
y = -1, 7

Question 2.
Solution:
Point A (0, 2) is equidistant from B (3, p) and also from C (p, 5)

Question 3.
Solution:
Three vertices of a rectangle ABCD are B (4, 0), C (4, 3) and D (0, 3)
Its diagonal are equal.

Question 4.
Solution:
Point P (k – 1, 2) is equidistant from two points A (3, k) and B (k, 5)

Question 5.
Solution:
Let P (x, 2) divides the join of A (12, 5) and B (4, -3) in the ration m : n.

Question 6.
Solution:

Question 7.
Solution:
Vertices of ∆ABC are A (7, -3), B (5, 3) and C (3, -1)

Question 8.
Solution:
Point C (k, 4) divides the join of A (2, 6) and B (5, 1) in the ratio 2 : 3

Question 9.
Solution:
Let a point P (x, 0) on x-axis is equidistant from two points A (-1, 0) and B (5, 0)

Question 10.
Solution:
Distance between two points

Question 11.
Solution:
The points (3, a) lies on the line 2x – 3y = 5
It will satisfy it.
2 x 3 – 3 x a = 5
6 – 3a = 5 => 3a = 6 – 5 = 1
a = \(\frac { 1 }{ 3 }\)

Question 12.
Solution:
Points A (4, 3) and B (x, 5) lie on the circle with centre O (2, 3)
OA = OB

Question 13.
Solution:
P (x, y) is equidistant from the point A (7, 1) and B (3, 5)

Question 14.
Solution:
O (0, 0) is the centroid of ∆ABC whose vertices are A (a, b), B (b, c) and C (c, a)

Question 15.
Solution:
Coordinates of centroid

Question 16.
Solution:
Let point P (4, 5) divides the join of A (2, 3) and B (7, 8) in the ratio m : n

Question 17.
Solution:
Points are given A (2, 3), B (4, k) and C (6, -3)
Points are collinear.

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry Ex 16C

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Question 1.
Solution:
(i) In ∆ABC, vertices are A (1, 2), B (-2, 3) and C (-3, -A)

Question 2.
Solution:
Vertices of quadrilateral ABCD are A (3,-1), B (9, -5), C (14, 0) and D (9, 19)
Join diagonal AC.

Question 3.
Solution:
PQRS is a quadrilateral whose vertices are P (-5, -3), Q (-4, -6), R (2, -3) and S (1, 2)
Join PR which forms two triangles PQR and PSR.

Question 4.
Solution:
ABCD is a quadrilateral whose vertices are A (-3, -1), B (-2, -4), C (4, -1) and D (3, 4)
Join AC which terms two triangles ABC and ADC

Question 5.
Solution:

Question 6.
Solution:
D, E and F are the midpoints of the sides CB, CA and AB respectively of ∆ABC.
Vertices are A (2, 1), B (4, 3) and C (2, 5)

Question 7.
Solution:
Vertices of ∆ABC are A (7, -3), B (5, 3) and C (3, -1)
D is midpoint of BC


Similarly area of ∆ACD = 5 sq. units (10 – 5 = 5 sq. units)
Hence, median divides the triangle into two triangles of equal in area.

Question 8.
Solution:
In ∆ABC, coordinates of A are (1, -4) and let C and E are the midpoints of AB and AC respectively.
Coordinates of F are (2, -1) and of E are (0, -1)
Let coordinates of B be (x1, y1) and of C be (x2, y2)

Question 9.
Solution:
A (6, 1), B (8, 2) and C (9, 4) are the three vertices of a parallelogram ABCD.
E is the midpoint of DC.
Join AE, AC and BD which intersects at O.
O is midpoint of AC

Question 10.
Solution:
(i) Vertices of a ∆ABC are A (1, -3), B (4, p) and C (-9, 7) and area = 15 sq. units

Question 11.
Solution:
Vertices of ∆ABC are A (k + 1, 1), B (4, -3) and C (7, -k) and area = 6 sq. units.

Question 12.
Solution:
Let the vertices of a triangle ABC are A (-2, 5), B (k, -4) and C (2k + 1, 10) and area = 53 sq. units

Question 13.
Solution:
(j) Points are A (2, -2), B (-3, 8) and C (-1, 4)

Question 14.
Solution:
Points are A (x, 2), B (-3, -4) and C (7, -5)

Question 15.
Solution:
Points are given A (-3, 12), B (7, 6) and C (x, 9)

Question 16.
Solution:
Points are given P (1, 4), Q (3, y) and R (-3, 16)

Question 17.
Solution:
The given points are A (-3, 9), B (2, y) and C (4, -5)

Question 18.
Solution:
The points are given A (8, 1), B (3, -2k) and C (k, -5)

Question 19.
Solution:
The points are given A (2, 1), B (x, y) and C (7, 5)

Question 20.
Solution:
The points are given A (x, y), B (-5, 7) and C (-4, 5)

Question 21.
Solution:
Points are given A (a, 0), B (0, b) and C (1,1)
Points are collinear.
Area of ∆ABC = 0

Question 22.
Solution:
The points are given P (-3, 9), Q (a, ti) and R (4, -5)
Points are collinear.
Area of ∆PQR = 0

Question 23.
Solution:
Vertices of ∆ABC are A (0, -1), B (2, 1) and C (0, 3)

Question 24.
Solution:
Let A (a, a²), B (b, b²) and C (0, 0)
For the points A, B and C to collinear area of ∆ABC must be zero.
Now, area of ∆ABC

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C 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.

RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry Ex 16B

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Question 1.
Solution:
(i) Let P (x, y) be the required point which divides the line joining the points A (-1, 7) and B (4, -3) in the ratio 2 : 3.

Question 2.
Solution:
Let (7, -2) and B (1, -5) be the given points and P (x, y) and Q (x’, y’) are the points of trisection.

Question 3.
Solution:
Let coordinates of P be (x, y) which divides the line segment A (-2, -2) and B (2, -4) in

Question 4.
Solution:
Let the coordinates of A be (x, y) which lies on line joining P (6, -6) and Q (-4, -1)

Question 5.
Solution:
Points P, Q, R and S divides a line segment joining the points A (1, 2) and B (6, 7) in 5 equal parts.

Question 6.
Solution:
Points P, Q and R in order divide a line segment joining the points A (1, 6) and B (5, -2) in 4 equal parts.

P divides AB in the ratio of 1 : 3 Let coordinates of P be (x, y), then

Question 7.
Solution:
The line segment joining the point A (3, -4) and B (1, 2) is trisected by the points P (p, -2) and Q(\(\frac { 1 }{ 2 }\), q).

Question 8.
Solution:
Mid point of the line segment joining the points A (3, 0) and B (-5, 4)

Question 9.
Solution:
(2, p) is the mid point of the line segment joining the points A (6, -5), B (-2, 11)

Question 10.
Solution:
Mid point of the line segment joining the points A (2a, 4) and B (-2, 3b) is C (1, 2a + 1)

Question 11.
Solution:
The line segment joining the points A (-2, 9) and B (6, 3) is a diameter of a circle with centre C.

Question 12.
Solution:
AB is diameter of a circle with centre C.
Coordinates of C (2, -3) and of B (1, 4)

Question 13.
Solution:
Let P (2, 5) divides the line segment joining the points A (8,2) and B (-6, 9) in the ratio m : n

Question 14.
Solution:

Question 15.
Solution:
Let P (m, 6) divides the join of A (-4, 3) and B (2, 8) in the ratio k : 1
Then coordinates of P will be

Question 16.
Solution:
Let point P (-3, k) divides the join of A (-5, -4) and B (-2, 3) in the ratio m : n, then

Question 17.
Solution:
Let point P on the x-axis divides the line segment joining the points A (2, -3) and B (5, 6) the ratio m : n
Let P is the point on x-axis whose coordinates are (x, 0)

Question 18.
Solution:
Let a point P on y-axis divides the line segment joining the points A (-2, -3) and B (3, 7) in the ratio m : n
Let the coordinates of P be (0, y)

Question 19.
Solution:
Let a point P (x, y) on the given line x – y – 2 = 0 divides the line segment joining the points A (3, -1) and B (8, 9) in the ratio m : n, then

Question 20.
Solution:
Vertices of ∆ABC are A (0, -1), B (2, 1) and C (0, 3)
Let AD, BE and CF are the medians of sides BC, CA and AB respectively, then
Coordinates of D will be =

Question 21.
Solution:
Centroid of ∆ABC where coordinates of A are (-1, 0), of B are (5, -2) and of C are (8, 2)

Question 22.
Solution:
G (-2, 1) is the centroid of ∆ABC whose vertex A is (1, -6) and B is (-5, 2)
Let the vertex C be (x, y), then

Question 23.
Solution:
O (0, 0) is the centroid of ∆ABC in which B is (-3, 1), C is (0, -2)
Let A be (x, y), then

Question 24.
Solution:
Points are A (3, 1), B (0, -2), C (1, 1) and D (4, 4)

Question 25.
Solution:
Points P (a, -11), Q (5, b), R (2, 15) and S (1, 1) are the vertices of a parallelogram PQRS.
Diagonals of a parallelogram bisect each other.
O is mid point of PR and QS.

Question 26.
Solution:
Three vertices of a parallelogram ABCD are A (1, -2), B (3, 6), C (5, 10).
Let fourth vertices D be (x, y)

Question 27.
Solution:
Let a point P (0, y) on 7-axis, divides the line segment joining the points (-4, 7) and (3, -7) in the ratio m : n

Question 28.
Solution:

Question 29.
Solution:
Let a point P (x, 0) divides the line segment joining the points A (3, -3) and B (-2, 2) in the ratio m : n

Question 30.
Solution:
Base QR of an equilateral APQR lies on x- axis is O (0, 0) is mid point PQR and coordinate of Q are (-4, 0).
Coordinate of R will be (4, 0)

Question 31.
Solution:
Base BC of an equilateral triangle ABC lies on y-axis in such a way that origin O (0, 0) lies is the middle of BC and coordinates of C are (0, -3).
Coordinates of B will be (0, 3)

Question 32.
Solution:
Let the points P (-1, y) lying on the line segment joining points A (-3, 10) and B (6, -8) divides it in the ratio m : n.

Question 33.
Solution:
In rectangle ABCD, A (-1, -1), B (-1, 4), C (5, 4), D (5, -1)
P, Q, R and S are the mid points of AB, BC, CD and DA respectively.

Question 34.
Solution:
P is mid point of line segment joining the points A (-10, 4) and B (-2, 0)

Question 35.
Solution:
Let coordinates of P and Q be (0, y) and (x, 0) respectively.
Let M (2, -5) be the mid-point of PQ.
By midpoint formula

Question 36.
Solution:

Question 37.
Solution:

Question 38.
Solution:
Let the other two vertices be (h, k) and (m, n).
Hence, the vertices in order are (3, 2), (-1, 0), (h, k) and (m, n).
It is to be kept in mind that the diagonals of a parallelogram bisect each other.
Hence, the point of intersection (2, -5) is the midpoint of the diagonal whose ends are (3, 2) and (h, k). Then

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B 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.

RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry Ex 16A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Question 1.
Solution:
(i) A (9, 3) and B (15, 11)

Question 2.
Solution:
Distance from origin O (0, 0) and the given points (x, y) = √(x² + y²)

Question 3.
Solution:
Points A (x, -1), B (5, 3) and AB = 5 units

Question 4.
Solution:
Points A (2, -3), B (10, y) and AB = 10

Question 5.
Solution:
Points P (x, 4), Q (9, 10) and PQ = 10

Question 6.
Solution:
Point A (x, 2) is equidistant from B (8, -2 and C (2, -2)

Question 7.
Solution:
A (0, 2) is equidistant from B (3, p) and C ip, 5)
Then AB = AC

Question 8.
Solution:
Let point P (x, 0) is on x-axis and P is equidistant from A (2, -5) and B (-2, 9)

Question 9.
Solution:
Let the points on x-axis be P (x,, 0) and Q (x2, 0) and A (11, -8)

or x – 5 = 0, then x = 5
Points are (17, 0) and (5, 0)

Question 10.
Solution:
Let point P (0, y) is on the y-axis, then

Question 11.
Solution:
P (x, y) is equidistant from A (5, 1) and B (-1, 5)

Question 12.
Solution:
P (x, y) is equidistant from A (6, -1) and B (2, 3)

Question 13.
Solution:
Let the coordinates of the points be O (x, y)

Question 14.
Solution:
Points A (4, 3) and B (x, 5) lie on a circle with centre O (2, 3)

Question 15.
Solution:
Point C (-2, 3) is equidistant from points A (3, -1) and B (x, 8)

Question 16.
Solution:
Point P (2, 2) is equidistant from the two points A (-2, k) and B (-2k, -3)

Question 17.
Solution:
(i) Let point P (x, y) is equidistant from A (a + b, b – a) and B (a – b, a + b), then

Question 18.
Solution:
We know that if the sum of any two of these distances is equal to the distance of the third, then the points are collinear.
Now, (i) Let the points are A (1, -1), B (5, 2), C (9, 5)

Question 19.
Solution:
The points are A (7, 10), B (-2, 5) apd C (3, -4)

Question 20.
Solution:
Points are A (3, 0), B (6, 4) and C (-1, 3)

Question 21.
Solution:
Points are A (5, 2), B (2, -2) and C (-2, t)

Question 22.
Solution:
Points are A (2, 4), B (2, 6) and C (2 + √3, 5)

Question 23.
Solution:
Let the points are A (-3, -3), B (3, 3), C (-3√3, 3√3)

Question 24.
Solution:
Points are A (-5, 6), B (3, 0), C (9, 8)

Question 25.
Solution:
Points are O (0, 0), A (3, √3) and B (3, -√3)

Question 26.
Solution:
(i) Points are A (3, 2), B (0, 5), C (-3, 2), D (0, -1)

Question 27.
Solution:
Points are A (-3, 2), B (-5, -5), C (2, -3), D (4, 4)

Question 28.
Solution:
Points are A (3, 0), B (4, 5), C (-1, 4) and D (-2, -1)

Question 29.
Solution:
Points are A (6, 1), B (8, 2), C (9, 4) and D (7, 3)

Question 30.
Solution:
Points are A (2, 1), B (5, 2), C (6, 4) and D (3, 3)

Question 31.
Solution:
Points are A (1, 2), B (4, 3), C (6, 6) and D (3, 5)

Question 32.
Solution:
(i) Points are A (-4, -1), B (-2, -4), C (4, 0) and D (2, 3)

Question 33.
Solution:

Question 34.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A 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.