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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?
Solution
`"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.
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