Class 10 Class 10 Maths RS Aggarwal Solutions Class 10

RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry

RS Aggarwal Solutions Class 10 Chapter 16
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RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry is a part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry.

RS Aggarwal Solutions Class 10 Chapter 16 Exercise 16A

Question 1:
(i) The given points are A(9,3) and B(15,11).
rs-aggarwal-class-10-solutions-co-ordinate-geometry-16-q1-1
(ii) The given points are A(7,4) and B(-5,1).
rs-aggarwal-class-10-solutions-co-ordinate-geometry-16-q1-2
(iii) The given points are A(-6, -4) and B(9,-12).
rs-aggarwal-class-10-solutions-co-ordinate-geometry-16-q1-3
(iv) The given points are A(1, -3) and B(4, -6).
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16a
(v) The given points are P(a + b, a – b) and Q(a – b, a + b).
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16a
(vi) The given points are P(a sin a, a cos a) and Q(a cos a, – a sina).
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16a

Question 2:
(i) The given point is A(5, -12) and let O(0,0) be the origin.
RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16a
(ii) The given point is B(-5, 5) and let O(0,0) be the origin.
RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16a
(iii) The given point is C(-4, -6) and let O(0,0) be the origin.
RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16a

Question 3:
The given points are A(a, -1) and B(5,3).
Solution of RS Aggarwal Class 10 Chapter 16 Co-ordinate Geometry ex 16a

Question 4:
Let R(10,y) be the point at a distance of 10 units from P(2, -3).
Solution of RS Aggarwal Class 10 Chapter 16 Co-ordinate Geometry ex 16a

Question 5:
Let A(6, -1), B(1,3) and C(k,8) are the given points.
Solution of RS Aggarwal Class 10 Chapter 16 Co-ordinate Geometry ex 16a

Question 6:
Let A(a, 2), B(8, -2) and C(2,-2) be the given points. Then first we find:
RS Aggarwal Solutions Class 10 2017 Chapter 16 Co-ordinate Geometry ex 16a
Therefore, a = 5

Question 7:
Let any point P on x – axis is (x,0) which is equidistant from A(-2, 5) and B(-2, 9).
RS Aggarwal Solutions Class 10 2017 Chapter 16 Co-ordinate Geometry ex 16a
This is not admissible.
Hence, there is no point on x – axis which is equidistant from A(-2, 5) and B(-2, 9).

Question 8:
Let any point P on x – axis is (0,y) which is equidistant from A(5, -2) and B(-3, 2)
RS Aggarwal Solutions Class 10 2017 Chapter 16 Co-ordinate Geometry ex 16a
Thus, the point on y – axis is (0, -2).

Question 9:
he point A(4,3) and B(x,5) lie on a circle. Its centre is O(2,3)
Class 10 RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16a

Question 10:
Let A(6, -1) and B(2,3) be the given point and P(x,y) be the required point, we get
Class 10 RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16a

Question 11:
Let A(11, -8) be the given point and let P(x,0) be the required point on x – axis
Then,
Class 10 RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16a
Hence, the required points are (17,0) and (5,0).

Question 12:
Let the required points be P(x,y), then
PA = PB = PC. The points A, B, C are (5,3), (5, -5) and (1, -5) respectively.
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 16 Co-ordinate Geometry ex 16a
Hence, the point P is (3, -1).

Question 13:
Let A(1, -1), B(5, 2) and C(9, 5) are the given points. Then
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 16 Co-ordinate Geometry ex 16a
Hence the given points A, B, C are collinear.

Question 14:
(i) Let A (6,9), B(0,1) and C(-6, -7) be the given points. Then
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 16 Co-ordinate Geometry ex 16a
Hence the given A, B, C are collinear.
(ii) Let A(-1, -1), B(2,3) and C(8,11) be the given points. Then
RS Aggarwal Class 10 Book pdf download Chapter 16 Co-ordinate Geometry ex 16a
Hence the given A, B, C are collinear.
(iii) Let P(1,1), Q(-2,7) and R(3, -3) be the given points, then
RS Aggarwal Class 10 Book pdf download Chapter 16 Co-ordinate Geometry ex 16a
(iv) Let P(2,0), Q(11,6) and R(-4,-4) be the given points
Then,
RS Aggarwal Class 10 Book pdf download Chapter 16 Co-ordinate Geometry ex 16a
Hence the given P, Q, R are collinear.

Question 15:
Let A(3,0), B(6,4) and C(-1,3) are the given points. Then
Class 10 Maths RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16a
∴ ∆ ABC is an isosceles right – angled triangle.
This shows that ∆ ABC is right angled at A.

Question 16:
Vertices of triangle ABC are A(7, 10), B(-2, 5) and C(3, -4)
Class 10 Maths RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16a
∴ ∆ ABC is a right angled triangle.
Hence ∆ ABC is an isosceles right triangle.

Question 17:
Let A(-5,6), B(3,0) and C(9,8) be the given points. Then
Class 10 Maths RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16a

Question 18:
Let O(0,0), A(3,√3) and B(3,-√3) are the given points.
Maths RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16a
Hence, DABC is equilateral and each of its sides being 2√3 units.
Maths RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16a

Question 19:
Let A(2,1), B(5,2), C(6,4) and D(3,3) are the angular points of a parallelogram ABCD. Then
Maths RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16a
Diagonal AC ≠ Diagonal BD.
Thus ABCD is not a rectangle but it is a parallelogram because its opposite sides are equal and diagonals are not equal.

Question 20:
(i) Let A(0, -4), B(6,2), C(3,5) and D(-3,-1) are the vertices of quad. ABCD. Then
RS Aggarwal Solutions Class 10 2018 Chapter 16 Co-ordinate Geometry ex 16a
Thus, ABCD is a quadrilateral whose opposite sides are equal and the diagonals are equal.
Hence, quad. ABCD is a rectangle.
(ii) Let A(2, -2), B(14, 10), C(11, 13) and D(-1, 1) be the angular points of quad. ABCD, then
RS Aggarwal Solutions Class 10 2018 Chapter 16 Co-ordinate Geometry ex 16a
Thus, ABCD is a quadrilateral whose opposite sides are equal and diagonals are equal.
Hence, quad. ABCD is rectangle.
(iii) Let A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3) are the vertices of quad. ABCD. Then
RS Aggarwal Solutions Class 10 2018 Chapter 16 Co-ordinate Geometry ex 16a
Thus, ABCD is a quadrilateral whose opposite sides are equal and the diagonals are equal
Hence, quad. ABCD is a rectangle.

Question 21:
(i) Let A(6,2), B(2,1), C(1,5) and D(5,6) be the angular points of quad. ABCD. Join AC and BD
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16a
Thus, ABCD is a quadrilateral in which all sides are equal and the diagonals are equal.
Hence, quad ABCD is a square.
(ii) Let P(0, -2), Q(3,1), R(0,4) and S(-3,1) be the angular points of quad. ABCD
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16a
Thus, PQRS is a quadrilateral in which all sides are equal and the diagonals are equal.
Hence, quad. PQRS is a square.

(iii) The angular points of quadrilateral ABCD are A(3,2), B(0,5), C(-3,2) and D(0,-1)
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16a
Thus, all sides of quad. ABCD are equal and diagonals are also equal.
Quad. ABCD is a square.

Question 22:
Let A(-3,2), B(-5, -5), C(2, -3) and D(4,4) be the angular point of quad ABCD. Join AC and BD.
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16a
Thus, ABCD is a quadrilateral having all sides equal but diagonals are unequal.
Hence, ABCD is a rhombus.
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16a

Exercise 16B

RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16b

Question 1:
The end points of AB are A(-1,7) and B(4, -3).
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16b
Let the required point be P(x, y)
By section formula, we have
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16b
Hence the required point is P(1, 3).

Question 2:
The end points of PQ are P(-5, 11) and Q(4, -7).
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16b
By section formula, we have
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16b
Hence the required point is (2, -3).

Question 3:
Let P(x, y) and Q(p,q) be the points of trisection of the line segment. Joining A(2,1) and B(5, -8)
The, P(x, y) divide AB in the ratio 1 : 2.
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16b
By section formula, we have
RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16b
Here m = 1 and n = 2
RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16b
P(3, -2) is the 1st point of trisection of AB
Also Q(p, q) divides AB in the ratio 2 : 1
Here m = 2 and n = 1
(x1=2, y1=1) and (x2=5, y2=-8)
RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16b
Q(4, -5) is the 2nd point of trisection of AB
Hence P(3, -2) and Q(4, -5) are the required point.

Question 4:
Let P(x, y) and Q(p, q) be the point of trisection of line segment A(-4, 0) and B(0, 6)
Then P(x, y) divides AB in the ratio 1 : 2
RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16b
Here m = 1 and n = 2
Solution of RS Aggarwal Class 10 Chapter 16 Co-ordinate Geometry ex 16b

Question 5:
Point P divides the join of A(3, -4) and B(1,2) in the ratio 1 : 2.
Coordinates of P are:
Solution of RS Aggarwal Class 10 Chapter 16 Co-ordinate Geometry ex 16b

Question 6:
Let (x, y) be the coordinates of a point P which divides the line joining A(4, -5) and B(4, 5) such that AP : AB = 2 : 5
Solution of RS Aggarwal Class 10 Chapter 16 Co-ordinate Geometry ex 16b
Coordinates of P are (4, -1)

Question 7:
(i) The coordinates of mid – points of the line segment joining A(3, 0) and B(-5, 4) are Solution of RS Aggarwal Class 10 Chapter 16 Co-ordinate Geometry ex 16b
(ii) Let M(x, y) be the mid – point of AB, where A is (-11, -8) and B is (8, -2). Then,
Solution of RS Aggarwal Class 10 Chapter 16 Co-ordinate Geometry ex 16b

Question 8:
The midpoint of line segment joining the points A(6, -5) and B(-2, 11) is
RS Aggarwal Solutions Class 10 2017 Chapter 16 Co-ordinate Geometry ex 16b
Also, given the midpoint of AB is (2, p)
⇒ p = 3

Question 9:
C(1, 2a + 1) is the midpoint of A(2a, 4) and B(-2, 3b)
RS Aggarwal Solutions Class 10 2017 Chapter 16 Co-ordinate Geometry ex 16b

Question 10:
Points P, Q, R divide the line segment joining the points A(1,6) and B(5, -2) into four equal parts
Point P divide AB in the ratio 1 : 3 where A(1, 6), B(5, -2)
RS Aggarwal Solutions Class 10 2017 Chapter 16 Co-ordinate Geometry ex 16b
Therefore, the point P is
RS Aggarwal Solutions Class 10 2017 Chapter 16 Co-ordinate Geometry ex 16b
Also, R is the midpoint of the line segment joining Q(3, 2) and B(5, -2)
RS Aggarwal Solutions Class 10 2017 Chapter 16 Co-ordinate Geometry ex 16b

Question 11:
Let A(-2, 9) and B(6, 3) be the two points of the given diameter AB and let C(a, b) be the center of the circle.
Then, clearly C is the midpoint of AB
Class 10 RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16b
By the midpoint formula of the co-ordinates,
Class 10 RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16b
Hence, the required point C(2, 6).

Question 12:
A, B are the end points of a diameter. Let the coordinates of A be (x, y).
The point B is (1, 4)
Class 10 RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16b
The center C(2, -3) is the midpoint of AB.
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 16 Co-ordinate Geometry ex 16b
The point A is (3, -10).

Question 13:
Let P divided the join of A(8, 2), B(-6, 9) in the ratio k : 1
By section formula, the coordinates of p are
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 16 Co-ordinate Geometry ex 16b
Hence, the required ratio of (\frac { 3 }{ 4 }   : 1) which is (3 : 4)

Question 14:
Let P(-6, a) divides the join of A(-3, -1) and B(-8, 9) in the ratio k : 1
Then the coordinates of P are given by
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 16 Co-ordinate Geometry ex 16b
But, we are given P is (-6, a)
RS Aggarwal Class 10 Book pdf download Chapter 16 Co-ordinate Geometry ex 16b
Hence, required ratio is 3 : 2 and a = 5

Question 15:
Let P divided the join of line segment A(-4, 3) and B(2, 8) in the ratio k : 1
∴ the point P is
RS Aggarwal Class 10 Book pdf download Chapter 16 Co-ordinate Geometry ex 16b

Question 16:
Let P is dividing the given segment joining A(-5, -4) and B(-2, 3) in the ratio r : 1
RS Aggarwal Class 10 Book pdf download Chapter 16 Co-ordinate Geometry ex 16b
Coordinates of point P
Class 10 Maths RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16b

Question 17:
Let the x- axis cut the join of A(2, -3) and B(5, 6) in the ratio k : 1 at the point P
Then, by the section formula, the coordinates of P are Class 10 Maths RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16b

Class 10 Maths RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16b
But P lies on the x axis so, its ordinate must be 0
Class 10 Maths RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16b
So the required ratio is 1 : 2
Thus the x – axis divides AB in the ratio 1 : 2
Putting rs-aggarwal-class-10-solutions-co-ordinate-geometry-16b-q17-4 we get the point P as
Maths RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16b
Thus, P is (3, 0) and k = 1 : 2

Question 18:
Let the y – axis cut the join A(-2, -3) and B(3, 7) at the point P in the ratio k : 1
Then, by section formula, the co-ordinates of P are
Maths RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16b
But P lies on the y-axis so, its abscissa is 0
Maths RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16b
So the required ratio is \frac { 2 }{ 3 }   : 1 which is 2 : 3
Putting Maths RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16b we get the point P as
rs-aggarwal-class-10-solutions-co-ordinate-geometry-16b-q18-4
i.e., P(0, 1)
Hence the point of intersection of AB and the y – axis is P(0, 1) and P divides AB in the ratio 2 : 3

Question 19:
Let the line segment joining A(3, -1) and B(8, 9) is divided by x – y – 2 = 0 in ratio k : 1 at p.
RS Aggarwal Solutions Class 10 2018 Chapter 16 Co-ordinate Geometry ex 16b
Coordinates of P are
RS Aggarwal Solutions Class 10 2018 Chapter 16 Co-ordinate Geometry ex 16b
Thus the line x – y – 2 = 0 divides AB in the ratio 2 : 3

Question 20:
Let D, E, F be the midpoint of the side BC, CA and AB respectively in ∆ABC.
RS Aggarwal Solutions Class 10 2018 Chapter 16 Co-ordinate Geometry ex 16b
Then, by the midpoint formula, we have
RS Aggarwal Solutions Class 10 2018 Chapter 16 Co-ordinate Geometry ex 16b
Hence the lengths of medians AD, BE and CF are given by
RS Aggarwal Solutions Class 10 2018 Chapter 16 Co-ordinate Geometry ex 16b

Question 21:
Here  RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16b
Let G(x, y) be the centroid of ∆ABC, then
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16b
Hence the centroid of ∆ABC is G(4, 0).

Question 22:
Two vertices of ∆ABC are A(1, -6) and B(-5, 2) let the third vertex be C(a, b)
Then, the co-ordinates of its centroid are
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16b
But given that the centroid is G(-2, 1)
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16b
Hence, the third vertex C of ∆ABC is (-2, 7).

Question 23:
Two vertices of ∆ABC are B(-3, 1) and C(0, -2) and third vertex be A(a, b)
Then the coordinates of its centroid are
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16b
Hence the third vertices A of ∆ABC is A(3, 1).

Question 24:
Let A(3,1), B(0, -2), C(1, 1) and D(4, 4) be the vertices of quadrilateral
Join AC, BD. AC and BD, intersect other at the point O.
RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16b
We know that the diagonals of a parallelogram bisect each other
Therefore, O is midpoint of AC as well as that of BD
Now midpoint of AC is RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16b

And midpoint of BD is RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16b
Mid point of AC is the same as midpoint of BD
Hence, A, B, C, D are the vertices of a parallelogram ABCD.

Question 25:
Let P(a, -11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a parallelogram PQRS.
RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16b
Join the diagonals PR and SQ.
They intersect each other at the point O. We know that the diagonals of a parallelogram bisect each other.
Therefore, O is the midpoint of PR as well as that of SQ
Now, midpoint of PR is RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16b
And midpoint of SQ is rs-aggarwal-class-10-solutions-co-ordinate-geometry-16b-q25-3
Solution of RS Aggarwal Class 10 Chapter 16 Co-ordinate Geometry ex 16b
Hence the required values are a = 4 and b = 3.

Question 26:
Let A(1, -2), B(3, 6) and C(5, 10) are the given vertices of the parallelogram ABCD.
Solution of RS Aggarwal Class 10 Chapter 16 Co-ordinate Geometry ex 16b
Let D(a, b) be its fourth vertex. Join AC and BD.
Let AC and BD intersect at the point O.
We know that the diagonals of a parallelogram bisect each other.
So, O is the midpoint AC as well as that of BD
Midpoint of AC is Solution of RS Aggarwal Class 10 Chapter 16 Co-ordinate Geometry ex 16b
Midpoint of BD is rs-aggarwal-class-10-solutions-co-ordinate-geometry-16b-q26-3
Solution of RS Aggarwal Class 10 Chapter 16 Co-ordinate Geometry ex 16b
Hence the fourth vertices is D(3, 2).

RS Aggarwal Solutions Class 10 Chapter 16 Exercise 16C

Question 1:
(i) Let A(1, 2), B(-2, 3) and C(-3, -4) be the vertices of the given ∆ ABC, then
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16c
(ii) The coordinates of vertices of ∆ ABC are A(-5, 7), B(-4, -5) and C(4, 5)
Here, x= -5, y1 = 7 ; x2 = -4, y2 = -5 ; x3 = 4, y3 = 5
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16c
(iii) The coordinates of ∆ ABC are A(3, 8), B(-4, 2) and C(5, -1)
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16c
(iv) Let P(10, -6), Q(2, 5) and R(-1, 3) be the vertices of the given ∆ PQR. Then,
rs-aggarwal-class-10-solutions-co-ordinate-geometry-16c-q1-4

Question 2:
(i) Join A and C, then area of quad. ABCD = area of ∆ ABC + area of ∆ ACD
RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16c
(ii) The vertices of quad. ABCD are A(0, 0), B(6, 0), C(4, 3) and D(0, 3)
RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16c
Area of quad. ABCD = Area of ∆ ABC + Area of ∆ ACD
= 9 + 6 = 15 sq. unit
(iii) Vertices of quad. ABCD are A(1, 0), B(5, 3), C(2, 7) and D(-2, 4)
RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16c
Vertices of ∆ABC are A(1, 0), B(5, 3), C(2, 7)
Solution of RS Aggarwal Class 10 Chapter 16 Co-ordinate Geometry ex 16c
Vertices of ∆ACD are A(1, 0), C(2, 7) and D(-2, 4)
Solution of RS Aggarwal Class 10 Chapter 16 Co-ordinate Geometry ex 16c
Area of quadrilateral ABCD
Solution of RS Aggarwal Class 10 Chapter 16 Co-ordinate Geometry ex 16c

Question 3:
(i) Let A(0, 1), B(1, 2) and C(-2, -1) be the given points. Then,
RS Aggarwal Solutions Class 10 2017 Chapter 16 Co-ordinate Geometry ex 16c
Hence the given points are collinear
(ii) Let A(-5, 1), B(5,5) and C(10, 7) be the given points.
RS Aggarwal Solutions Class 10 2017 Chapter 16 Co-ordinate Geometry ex 16c
(iii) Let P(a, b + c), Q(b, c + a) and R(c, a + B) be the given points.
RS Aggarwal Solutions Class 10 2017 Chapter 16 Co-ordinate Geometry ex 16c

Question 4:
(i) The given points are A(-1, 3), B(2, p) and C(5, -1)
Class 10 RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16c
(ii) The given points are A(3, 2), B(4, p) and C(5, 3)
Class 10 RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16c
(iii) The three points are A(-3, 9), B(2, p), C(4, -5)
Class 10 RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16c

Question 5:
The given points are A(-3, 12), B(7, 6) and C(x, 9)
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 16 Co-ordinate Geometry ex 16c

Question 6:
Let P(1, 4), Q(3, y) and R(-3, 16)
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 16 Co-ordinate Geometry ex 16c

Question 7:
The given points are A(x, y), B(-5, 7) and C9-4, 5)
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 16 Co-ordinate Geometry ex 16c
The given points A, B, C are collinear
RS Aggarwal Class 10 Book pdf download Chapter 16 Co-ordinate Geometry ex 16c

Question 8:
The vertices of a quadrilateral ABCD are (-4, -2), B(-3, -5), C(3, -2) and D(2, k)
Join AC.
RS Aggarwal Class 10 Book pdf download Chapter 16 Co-ordinate Geometry ex 16c
Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD
Now area of ∆ABC
RS Aggarwal Class 10 Book pdf download Chapter 16 Co-ordinate Geometry ex 16c

Question 9:
The vertices of ABC are A(4, -6), B(3, -2), C(5, 2)
AD is the median.
D is the midpoint of BC.
the coordinates of point D are rs-aggarwal-class-10-solutions-co-ordinate-geometry-16c-q9-1
Class 10 Maths RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16c
Vertices of ABD are A(4, -6), B(3, -2), D(4, 0)
Area of ∆ABD
Class 10 Maths RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16c

Question 10:
Vertices of ∆ABC are A(2, 1), B(x, y) and C(7, 5)
Class 10 Maths RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16c
The points A, B and C are collinear
area of ∆ABC = 0
Or 4x – 5y – 3 = 0

Question 11:
The vertices of ∆ABC are (a, 0), (0, b), C(1, 1)
Maths RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16c
The points A, B, C are collinear
Area of ∆ABC = 0
ab – a – b = 0 a + b = ab
Dividing by ab
RS Aggarwal Solutions Class 10 2018 Chapter 16 Co-ordinate Geometry ex 16c

Exercise 16D

Question 1:
Distance between the points rs-aggarwal-class-10-solutions-co-ordinate-geometry-16d-q1-1
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry ex 16d

Question 2:
The points (3, a) lies on the line 2x – 3y = 5.
Substituting the values of x and y in the given equation:
2 × 3 – 3 × a = 5 or 6 – 3a = 5
⇒ 3a = 1
⇒ a = \frac { 1 }{ 3 }

Question 3:
The points A(4,3) and B(x, 5) lie on the circle with center O(2,3)
OA and OB are radius of the circle.
RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry ex 16d

Question 4:
The point P(x, y) is equidistant from point A(7, 1) and B(3, 5)
Solution of RS Aggarwal Class 10 Chapter 16 Co-ordinate Geometry ex 16d

Question 5:
The vertices of ∆ABC are (a, b), (b, c) and (c, a)
Centroid is RS Aggarwal Solutions Class 10 2017 Chapter 16 Co-ordinate Geometry ex 16d
But centroid is (0, 0)
⇒ a + b + c = 0

Question 6:
The vertices of ∆ABC are A(2, 2), B(-4, -4) and C(5, -8)
The centroid of ∆ABC is given by

Class 10 RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16d

Question 7:
Let the point C(4, 5) divides the join of A(2, 3) and B(7, 8) in the ratio k : 1
The point C is RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 16 Co-ordinate Geometry ex 16d
But C is (4, 5)
RS Aggarwal Class 10 Book pdf download Chapter 16 Co-ordinate Geometry ex 16d
Thus, C divides AB in the ratio 2 : 3

Question 8:
The points A(2, 3), B(4, k) and C(6, -3) are collinear if area of ∆ABC is zero
Class 10 Maths RS Aggarwal Solutions Chapter 16 Co-ordinate Geometry ex 16d
But area of ABC = 0,
⇒ k = 0

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RS Aggarwal Solutions Class 10

  1. Real Numbers
  2. Polynomials
  3. Linear equations in two variables
  4. Triangles
  5. Trigonometric Ratios
  6. T-Ratios of Some Particular Angles
  7. Trigonometric Identities
  8. Trigonometric Ratios of Complementary Angles
  9. Mean, Median, Mode of Grouped Data, of Grouped Data Ex.9A Ex.9B Ex.9C and 9D Ex.9E
  10. Quadratic Equations Ex.10A Ex.10B Ex.10C Ex.10D
  11. Arithmetic Progressions
  12. Circles
  13. Constructions
  14. Height and Distance
  15. Probability
  16. Coordinate Geometry
  17. Perimeter and Areas of Plane Figures
  18. Areas of Circle, Sector and Segment
  19. Volume and Surface Areas of Solids

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