Find the value of k so that the area of the triangle with vertices A(k+1, 1), B(4, -3) and C(7, -k) is 6 square units.

Find the values of k so that the area of the triangle with vertices (1, -1), (-4, 2k) and (-k, -5) is 24 sq. units.

What type of triangle is DeltaABC , with the coordinates of the vertices are A(-1, 0) B(1, 0) and C(0, sqrt(3)) . Calculate its Area. OR Find the values of k so that the area of the triangle with vertices (1, -1) (-4, 2k) and (-k, -5) is 24 sq. units.

Find the value of k so that the area of the triangle with vertices (1,-1),(-4,2k)and (-k,-5) is 24 square units.

Find the values of k so that the area of the triangle with vertices (1,-1),(-4,2k) and (c ,-5 is 24sq .units.

Using determinats, find the values of k, if the area of triangle with vertices (-2, 0), (0, 4) and (0, k) is 4 square units.

Using determinant find the values of k so that the area of the triangle with vertices at the points (k,0),(4,0) and (0,2) is 4 sq units.