A pair of lines drawn through the origin forms a right angled isosceles triangle with right angle at the…
A pair of lines drawn through the origin forms a right angled isosceles triangle with right angle at the origin with the line $2 x+3 y=6$. The area (in sq. units) of the triangle thus formed is
$\frac{36}{13}$
$\frac{32}{13}$
$\frac{18}{5}$
$\frac{25}{9}$
Solution
Any line through the origin making an angle of $45^{\circ}$ with the given line $2 x+3 y=6$ is of the form $y=m x$ where
$\begin{aligned} & \tan \left( \pm 45^{\circ}\right)=\frac{m-\left(\frac{-2}{3}\right)}{1+m\left(\frac{-2}{3}\right)}= \pm 1 \\ & \therefore 3 m+2= \pm(3-2 m) \\ & \Rightarrow m=\frac{1}{5},-5\end{aligned}$
Hence the sides are $x-5 y=0,5 x+y=0$ and $2 x+3 y=6$ If $p$ be perpendicular from vertex $A(0,0)$ to base, then $p=\frac{6}{\sqrt{13}}$. Hence its area $=\frac{1}{2} A L \cdot B C$
$=\frac{1}{2}(p)(2 p)=p^2=\frac{36}{13}$ sq. units.
