Question

The vertex of a right angled triangle lies on the straight line $2x+y-10=0$ and the two of 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)

$3$

$2x+y-10=0$

$y=10-2x$

$\therefore (x,10-2x)$ is the points of the 3rd vertex

$(2,-3),(x+10-2x),(4,1)$

$\left(\frac{10-2x-1}{x-4}\right)\left(\frac{10-2x+3}{x-2}\right)=-1$

$(9-2x)(13-2x)=-(x-2)(x-4)$

$117-18x-26x+4x^2=-x^2+6x-8$

$5x^2-50x+125=0$

Solving the above quadratic equation, we get,

$x=5$

$y=10-2x\Rightarrow y=0$

The 3rd vertex is $(5,0)$

$A=\frac{1}{2}bh$

$=\frac{1}{2}\sqrt{(4-5{)}^2+(1-0{)}^2}\sqrt{(5-2{)}^2+(0+3{)}^2}$

$=\frac{1}{2}\sqrt{2}\times 3\sqrt{2}$

$=3$sq. units