Question

The area of the triangle whose two of the vertices are A(0, 1), B(3, 4) and the intersection point of median from A with side BC is (–2, 3), is (in square units)

Answer 1

Let D (-2, 3) is point of intersection of median from A with side BC.

∴ BD = DC

or D is midpoint of line BC.

Let point c be (x, y).

∴ -2 \(=\frac{x+3}{2}\) ⇒ x = -4 - 3 = -7

and 3 \(=\frac{y+4}{2}\) ⇒ y = 6 - 4 = 2

Third vertex of triangle is c(-7, 2)

Area of triangle is \(=|\frac12\) (0(2-4) + 3(1-2) -7 (4-1))|

\(=|\frac12\) (3 x -1 - 7 x 3)|

\(=|\frac12\) x -24| = |-12| = 12 square units.