Question

Find the area of triangle $△ABC$ having vertices $A(4,2),B(3,9)$ and $C(10,10)$.

Answer 1

$△ABC$ has vertices $A(4,2),B(3,9)$ and $C(10,10)$.
The area of a triangle formed by joining the points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is
$\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$
$\therefore$ Area of $△ABC$
$\frac{1}{2}\left|4\right(9-10)+3(10-2)+10(2-9\left)\right|$
$=\frac{1}{2}|-4+24-70|=\frac{50}{2}=25$

Hence, area of triangle $ABC$ is $25$.