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 square units?

Answer 1

`"Let " A(x_1=-2,y_1=5),B(x_2=k,y_2=-4) and C (x_3 = 2k+1,y_3=10)` be the vertices of the  triangle, So

`"Area " (ΔABC) =1/2 [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]`

`⇒53=1/2[(-2)(-4-10)+k(10-5)+(2k+1)(5+4)]`

`⇒53=1/2[28+5k+9(2k+1)]`

⇒ 28+5k+18k+9=106

⇒37+23k=106

⇒23k=106-37=69

`⇒k=69/23=3`

Hence , k=3.