The vertices of △ ABC are -2 1 5 4 and 2 -3 respectivel

SOLUTION

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

$=\frac{1}{2}|-2(4+3)+5(-3-1)+2(1-4)|$

$=20sq.units$

Now, $Area=\frac{1}{2}\times Base\times Altitude$

$\Rightarrow 20=\frac{1}{2}\times BC\times Altitude$

$\Rightarrow 20=\frac{1}{2}\times \sqrt{(5-2{)}^2+(4+3{)}^2}\times Altitude$

$\Rightarrow 20=\frac{1}{2}\times \sqrt{58}\times Altitude$

$\therefore Altitude=\frac{40}{\sqrt{58}}$