If area of triangle is 35 vuits with Vertices A(2-6), B(5,4) and C(K, 4) then K is

If area of triangle is 35 sq units with vertices (2, - 6), (5, 4) a n d (k , 4) . Then k is(A) 12 (B) -2 (C) 12 , 2 (D) 12 , -2

Using determinant, find k if area of triangle is 35 sq. units with vertices (2, -6), (5, 4) and (k, 4). a) 12 b) 2 c) -12,-2 d) 12,-2

area of the triangle with vertices A(3,4,-1), B(2,2,1) and C(3,4,-3) is :

Using vectors, find the area of triangle with vertices A(2,3,5), B(3,5, 8) and C(2, 7, 8) .

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.

Find the value of x if the area of triangle is 35 sq. units whose vertices are (x,4),(2,-6),(5,4).

Find the area of the triangle with vertices A\ (5,\ 4),\ \ B(-2,\ 4) and C\ (2,\ -6) .

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 area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

The area of a triangle with vertices (a,b+c), (b,c+a) and (c,a+b) is