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Question
Find the values of k so that the area of the triangle with vertices (k + 1, 1), (4, -3) and (7, -k) is 6 sq. units.
Solution
Let A(k + 1, 1), B(4,-3) and c(7,-k) are the vertices of the triangle.
Given that the area of the triangle is 6sq. units.
Area of the triangle is given by
`A=1/2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]`
`6=1/2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]`
`12=(k+1)(-3+k)+4(-k-1)+7(1+3)`
12=−3k+
`12=-3k+k^2-3+k-4k-4+28`
`12=k^2-6k+21`
`k^2-6k+21-12=0`
`k^2-6k+9=0`
`(k-3)^2=0`
`k=3,3`
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