. Find k if the area of the triangle with vertices at A(k, 2), B(-2, 1), C(1,-3) is sq. units

The area of a triangle with vertices at (x1, y1), (x2, y2), and (x3, y3) is given by the formula:

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

In this case, the vertices are A(k, 2), B(-2, 1), and C(1, -3). So, x1 = k, y1 = 2, x2 = -2, y2 = 1, x3 = 1, and y3 = -3.

Substitute these values into the formula:

Area = 1/2 * |k(1 - (-3)) + (-2)(-3 - 2) + 1(2 - 1)|

Simplify the equation:

Area = 1/2 * |4k + 10 + 1|

Area = 1/2 * |4k + 11|

Given that the area is 5 square units, we can set up the equation:

5 = 1/2 * |4k + 11|

Multiply both sides by 2 to get rid of the fraction:

10 = |4k + 11|

This absolute value equation has two possible solutions:

4k + 11 = 10 or 4k + 11 = -10

Solving these equations gives:

k = -1/4 or k = -21/4

So, the possible values for k are -1/4 or -21/4.