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

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 16 Co-ordinate Geometry Ex 16 B Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 16 Co-ordinate Geometry Ex 16 B Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16 B Maths book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 16
Chapter NameCoordinate Geometry
Exercise16 B
CategoryRS Aggarwal Solutions

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

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 16b 1.1
Let the required point be P(x, y)
By section formula, we have
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 1.2
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 16b 2.1
By section formula, we have
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 2.2
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 16b 3.1
By section formula, we have
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 3.2
Here m = 1 and n = 2
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 3.3
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 Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 3.4
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 Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 4.1
Here m = 1 and n = 2
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 4.2

Question 5:
Point P divides the join of A(3, -4) and B(1,2) in the ratio 1 : 2.
Coordinates of P are:
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 5.1

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
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 6.1
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 RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Q7
(ii) Let M(x, y) be the mid – point of AB, where A is (-11, -8) and B is (8, -2). Then,
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 7.1

Question 8:
The midpoint of line segment joining the points A(6, -5) and B(-2, 11) is
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 8.1
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 Chapter 16 Co-ordinate Geometry 16b 9.1

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 Chapter 16 Co-ordinate Geometry 16b 10.1
Therefore, the point P is
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 10.2
Also, R is the midpoint of the line segment joining Q(3, 2) and B(5, -2)
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 10.3

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
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 11.1
By the midpoint formula of the co-ordinates,
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 11.2
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)
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 12.1
The center C(2, -3) is the midpoint of AB.
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 12.2
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 Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 13.1
Hence, the required ratio of (34 : 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 Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 14.1
But, we are given P is (-6, a)
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 14.2
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 Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 15.1

Question 16:
Let P is dividing the given segment joining A(-5, -4) and B(-2, 3) in the ratio r : 1
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 16.1
Coordinates of point P
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 16.2

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 RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 17.1

RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 17.2
But P lies on the x axis so, its ordinate must be 0
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 17.3
So the required ratio is 1 : 2
Thus the x – axis divides AB in the ratio 1 : 2
Putting RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 17.4 we get the point P as
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 17.5
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
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 18.1
But P lies on the y-axis so, its abscissa is 0
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 18.2
So the required ratio is 23 : 1 which is 2 : 3
Putting RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 18.3 we get the point P as
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 18.4
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 Chapter 16 Co-ordinate Geometry 16b 19.1
Coordinates of P are
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 19.2
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 Chapter 16 Co-ordinate Geometry 16b 20.1
Then, by the midpoint formula, we have
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 20.2
Hence the lengths of medians AD, BE and CF are given by
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 20.3

Question 21:
Here RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 21.1
Let G(x, y) be the centroid of ∆ABC, then
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 21.2
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 16b 22.1
But given that the centroid is G(-2, 1)
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 22.2
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 16b 23.1
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 Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 24.1
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 Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 24.2

And midpoint of BD is RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 24.3
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 Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 25.1
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 Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 25.2
And midpoint of SQ is RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 25.3
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 25.4
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.
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 26.1
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 RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 26.2
Midpoint of BD is RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 26.3
RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry 16b 26.4
Hence the fourth vertices is D(3, 2).

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