Math, asked by pratikkumbhar004, 6 months ago

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

Answers

Answered by kartik2507
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

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