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

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 16 Co-ordinate Geometry Ex 16 C 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 C Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16 C Maths book pdf download. Now you will get step by step solution to each question.

### RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate GeometryEx 16 C

Question 1:
(i) Let A(1, 2), B(-2, 3) and C(-3, -4) be the vertices of the given ∆ ABC, then

(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

(iii) The coordinates of ∆ ABC are A(3, 8), B(-4, 2) and C(5, -1)

(iv) Let P(10, -6), Q(2, 5) and R(-1, 3) be the vertices of the given ∆ PQR. Then,

Question 2:
(i) Join A and C, then area of quad. ABCD = area of ∆ ABC + area of ∆ ACD

(ii) The vertices of quad. ABCD are A(0, 0), B(6, 0), C(4, 3) and D(0, 3)

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)

Vertices of ∆ABC are A(1, 0), B(5, 3), C(2, 7)

Vertices of ∆ACD are A(1, 0), C(2, 7) and D(-2, 4)

Question 3:
(i) Let A(0, 1), B(1, 2) and C(-2, -1) be the given points. Then,

Hence the given points are collinear
(ii) Let A(-5, 1), B(5,5) and C(10, 7) be the given points.

(iii) Let P(a, b + c), Q(b, c + a) and R(c, a + B) be the given points.

Question 4:
(i) The given points are A(-1, 3), B(2, p) and C(5, -1)

(ii) The given points are A(3, 2), B(4, p) and C(5, 3)

(iii) The three points are A(-3, 9), B(2, p), C(4, -5)

Question 5:
The given points are A(-3, 12), B(7, 6) and C(x, 9)

Question 6:
Let P(1, 4), Q(3, y) and R(-3, 16)

Question 7:
The given points are A(x, y), B(-5, 7) and C9-4, 5)

The given points A, B, C are collinear

Question 8:
The vertices of a quadrilateral ABCD are (-4, -2), B(-3, -5), C(3, -2) and D(2, k)
Join AC.

Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD
Now area of ∆ABC

Question 9:
The vertices of ABC are A(4, -6), B(3, -2), C(5, 2)
D is the midpoint of BC.
the coordinates of point D are

Vertices of ABD are A(4, -6), B(3, -2), D(4, 0)
Area of ∆ABD

Question 10:
Vertices of ∆ABC are A(2, 1), B(x, y) and C(7, 5)

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)

The points A, B, C are collinear
Area of ∆ABC = 0
ab – a – b = 0 a + b = ab
Dividing by ab

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