Q2 of 112 Page 150

Vertices of the triangles taken in order and their areas are given below. In each of the following find the value of a.

Vertices: (a, a), (4, 5), (6, –1)


Area (in sq. units): 9

Vertices of triangle A (a, a), B (4, 5) and C (6, –1)


Area of triangle = 9 sq. units


Area of triangle =


x1 = a, x2 = 4 and x3 = 6


y1 = a, y2 = 5 and y3 = –1


⇒


⇒


⇒


⇒


⇒ 9 × 2 = 8a – 34


⇒ 18 = 8a – 34


⇒ 8a = 18 +34


⇒ 8a = 52


⇒


⇒


Therefore, the required vertices are


More from this chapter

All 112 →
1

Find the area of the triangle formed by the points

(–4, –5), (4, 5) and (–1, –6)

 2

Vertices of the triangles taken in order and their areas are given below. In each of the following find the value of a.

Vertices: (0, 0), (4, a), (6, 4)


Area (in sq. units): 17

 2

Vertices of the triangles taken in order and their areas are given below. In each of the following find the value of a.

Vertices: (a, –3), (3, a), (–1,5)


Area (in sq. units): 12

 3

Determine if the following set of points are collinear or not.

(4, 3), (1, 2) and (–2, 1)