Question

Find the area of triangle whose vertices are (1,3), (-3,-5). 12,-1)​

Answer 1

Step-by-step explanation:

area of triangle= (1/2) [x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3(y1 – y2)]

A = (1, 3). B = (-3, -5) C = (12, -1)

(x1, y1). (x2, y2) (x3, y3)

$= \frac{1}{2} |1( - 5 - ( - 1) + ( - 3)( - 1 - 3) + 12(3 - ( - 5)| \\ \\ = \frac{1}{2} |1 ( - 5 + 1) + ( - 3)( - 4) + 12(3 + 5)| \\ = \frac{1}{2} |(1 \times - 4) + ( - 3)( - 4) + (12 \times 8)| \\ = \frac{1}{2} | - 4 + 12 + 96| \\ = \frac{1}{2} |104| \\ = \frac{1}{2} \times 104 \\ = 52 \: sq \: units$

Therefore the area of triangle = 52 sq units

Hope you get your answer