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  4. The vertex of a right angle of a right angled triangle lies on the straight line 2 x+y-10=0 and the two other vertices, at point (2,-3) and (4,1) then the area of triangle in sq. units is

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Q. The vertex of a right angle of a right angled triangle lies on the straight line $2 x+y-10=0$ and the two other vertices, at point $(2,-3)$ and $(4,1)$ then the area of triangle in sq. units is

Straight Lines

Solution:

$M _{1} \,M _{2}=-1 ; $
$\frac{9-2 a }{ a -4} \times \frac{13-2 a }{ a -2}=-1$
image
$117-26 a-18 a+4 a^{2}$
$=-\left(a^{2}-6 a+8\right) $
$5 a^{2}-50 a+125=0 ; $
$a=5$ so $ B$ is $(5,0)$
So area $=\frac{1}{2} AB \times AC $
$=\frac{1}{2} \sqrt{2} \times 3 \sqrt{2}=3$