#### 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`

Is there an error in this question or solution?

Solution 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. Concept: Area of a Triangle.