
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Last updated at Aug. 18, 2021 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Ex 4.3,1 Find area of the triangle with vertices at the point given in each of the following: (iii) (−2, −3), (3, 2), (−1, −8) The area of triangle is given by ∆ = 12 x1y11x2y21x3y31 Here, x1 = – 2, y1 = – 3 x2 = 3 , y2 = 2 x3 = – 1, y3 = – 8 ∆ = 12 −2−31321−1−81 = 12 −2 21−81−(−3) 31−11+1 32−1−8 = 12 [ – 2(2 – ( – 8)) + 3(3 – ( – 1)) + 1 ( – 24 – ( –2) ] = 12 [ – 2(2 + 8) + 3(3 + 1) + 1 (– 24 + 2) ] = 12 [ – 2(10) + 3 (4) + 1 ( – 22) ] = 12 ( – 20 – 22 + 12) = 12 (− 30) = – 15 Since the area of triangle is always positive, the required area of triangle is 15 square units