Question

For what value of k, (k> 0), is the area of the triangle with vertices (-2, 5), (k, -4) and (2k + 1, 10) equal to 53 sq. units?​

Answer 1

Answer:

Step-by-step explanation

Given Vertices are  (-2, 5), (k, -4) and (2k + 1, 10)

Area is 52 sq.units

Area of Triangle=1/2*x1(y2-y3)+x2(y3-y2)+x3(y1-y3)

                         53 = 1/2*-2(-4-10)+k(10-(-4))+2k+1(5-10)

                         53 =1/2*+28+14k-10k-5

                         106=23+4k

                         4k=53

                           k=13.2

 

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