The coordinates of the vertices of `ABC` are `A(4,1), B(-3,2)` and `C(0,k).` Given that the area of `ABC` is `12` unit, find the value of `k`

The co-ordinates of the vertices of Delta ABC are A(4,1),B (-3,2) and C(o,k).Given area of Delta ABC =12 square unit.Find the value of k.

If the coordinates of the vertices A,B,C of the ABC be (-4,2),(12,-2) and (8,6) respectively,then /_B=

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.

For a DeltaABC the vertices are A(1, 3), B(0, 0) and C(k, 0) . area of △ABC is 3. Find the value of k.

Let A(1, 3), B(0, 0) and C(k, 0) be three points such that ar(triangle ABC) =3 sq units. Find the value of k.

The vertices of a triangle are (1,-3), (4,k) and (-9,7) . If its area is 15 sq units, find the value of k.

If the vertices of a triangle ABC are A (-4, -1), B (1,2) and C (4, -3), then the coordinates of the circumcentre of the triangle are,

If the coordinates of the midpoints of the sides of Delta ABC are (1,1), (2,-3) and (3,4) , find the coordinates of the vertices of Delta ABC