Question

vertices of given triangles are taken in order and their areas are provided aside.in each case find the value of'p' ii) vertices-(p,p),(5,6),(5,-2) area (sq.units ) - 32 .​

Answer 1

Answer:

p = 1

Step-by-step explanation:

area of triangle= x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)

-32 = p{6-(-2)} + 5(-2-p) + 5(p-6)

-32 = p(6+2) + 5(-2-p) + 5(p-6)

-32 = p(8) + 5(-2-p) + 5(p-6)

-32 = 8p -10 - 5p + 5p - 30

-32 = 8p - 40

8p = -32 + 40

8p = 8

p = 1

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