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, –3), (3, a), (–1,5)


Area (in sq. units): 12

Vertices of triangle A (a, –3), B (3, a) and C (–1, 5)


Area of triangle = 12 sq. units


Area of triangle =


x1 = a, x2 = 3 and x3 = –1


y1 = –3, y2 = a and y3 = 5


⇒


⇒


⇒


⇒ 12 × 2 = a2 – 4a + 27


⇒24 = a2 – 4a + 27


⇒ a2 – 4a + 27 – 24 = 0


⇒ a2 – 4a + 3


⇒ a2 – 3a – a + 3 = 0


⇒ a (a – 3) – (a – 3) = 0


⇒ (a – 3) (a – 1) = 0


a – 3 = 0 or a – 1 = 0


a = 3 or a = 1


Therefore, the required vertices are (3, –3) or (1, –3)


More from this chapter

All 112 →
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, a), (4, 5), (6, –1)


Area (in sq. units): 9

 3

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

(4, 3), (1, 2) and (–2, 1)

 3

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

(–2, –2), (–6, –2) and (–2, 2)