Question

The area of triangle with vertices (2, 0), (6, 0) and (1, 4) is :​

Answer 1

Answer:

$x = 2 \\ x2 = 6 \\ x3 = 1 \\ \\ y1 = 0 \\ y2 = 0 \\ y3 = 4 \\ \\ \frac{1}{2} \times (2(0 + 0)6(0 + 4)1(0 + 0)) \\ = > \frac{1}{2} \times (2(0) + 6(4) + 1(0)) \\ = > \frac{1}{2} \times (0 + 16 + 0) \\ = > \frac{1}{2} \times 16 \\ = > 8sq.units$