Question

Find the area of triangle formed by the points A(5,2), B(4,7) and C(7,-4).

__ square units

Answer 1

If $A(x_1,y_1)$, $B(x_2,y_2)$ , $C(x_3,y_3)$ are vertices of triangle ABC, then its area is equal to

$△ABC=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$

Here, vertices are placed in anticlockwise if vertices are placed in the clockwise sense =, the above formula given negative value.

After simplification we get,

Area of triangle = $\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

Substituting the value of $(x_1,y_1)(x_2,y_2)$ and $(x_3,y_3)$.

= $\frac{1}{2}\left[5\right(7+4)+4(-4-2)+7(2-7\left)\right]$

= $\frac{1}{2}[55-24-35]=-\frac{4}{2}=-2$

Since area is a measure, which cannot be negative, we will take the numerical value of -2

Therefore, the area the triangle is 2 square units.