Question

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

• A. $\sqrt{10}$
• B. $3$
• C. $\frac{33}{5}$
• D. $11$

Answer 1

Answer: (B)

$3$

Line $A$ and $B$ have the gradient of 2.

So the gradient of the line vertical to them is $-\frac{1}{2}$

This its equation is, $y=-\frac{1}{2}x+c$

It passes through the point $(4,1)$ giving,$1=-\frac{1}{2}\left(4\right)+c⟹c=3$

Thus we have:

$y=-\frac{1}{2}x+3⟹x=-2y+6-\left(1\right)$

$y=2x-10-\left(2\right)$

solving (1) and (2), we get,

$y=\frac{2}{5}$ and $x=\frac{26}{5}$

​

verticle height between $(4,1),(\frac{26}{5},\frac{2}{5})$ is given by,

Area$=\frac{1}{2}\left(base\right)\left(height\right)$

$=\frac{1}{2}\left(\sqrt{(4-2{)}^2+(1+3{)}^2}\right)\left(\sqrt{(\frac{26}{5}-4{)}^2+(\frac{2}{5}-1{)}^2}\right)$

$=\frac{1}{2}\left(2\sqrt{5}\right)\left(\frac{3}{\sqrt{5}}\right)$

$=3sq.units$