Question

A pair of straight lines drawn through the origin forms an isosceles right triangle with the line $2x+3y=6$, then the lines and the area of the triangle thus formed is

• A. $\begin{array}{c}x-5y=0\\ 5x+y=0\\ \Delta =\frac{36}{13}\end{array}$
• B. $\begin{array}{c}3x-y=0\\ x+3y=0\\ \Delta =\frac{12}{17}\end{array}$
• C. $\begin{array}{c}5x-y=0\\ x+5y=0\\ \Delta =\frac{13}{5}\end{array}$
• D. None of these

Answer 1

The correct option is A $\begin{array}{c}x-5y=0\\ 5x+y=0\\ \Delta =\frac{36}{13}\end{array}$

OA and OB pass through origin,

$\therefore yintercept=0$

$\therefore y=m1x\left(OB\right)$

$y=m2x\left(OA\right)$

$Bothmake{45}^{\circ }withAB$

($\because$Triangle is right angled isoscles)

$2x+3y=6$

$⟹y=\frac{-2x}{3}+\frac{6}{2}$

$y=\frac{-2x}{3}+3$

$\tan \left(45\right)=\left|\frac{m1-\left(\frac{-2}{3}\right)}{1-\frac{2}{3}m1}\right|$

$Case1:$

$⟹1=\frac{m+\frac{2}{3}}{1-\frac{2}{3}m}$

$⟹1-\frac{2}{3}m=m+\frac{2}{3}$

$⟹\frac{1}{3}=\frac{5m}{3}$

$⟹m=\frac{1}{5}$

$Case:2$

$1=\frac{-(m+\frac{2}{3})}{1-\frac{2m}{3}}$

$⟹-1+2m=m+\frac{2}{3}$

$⟹-1-\frac{2}{3}=m-\frac{2m}{3}$

$⟹\frac{m}{3}=\frac{-5}{3}$

$⟹m=-5$

$\therefore m1=\frac{1}{5}$

$andm2=-5$

$\therefore y=\frac{x}{5}$

$⟹5y-x=0$

$andy=-5x$

$⟹5x+y=0$

are the equations of two lines.

To find points A and B

$AB:2x+3y=6$

$OA:y=-5x$

$⟹ForA:$

$2x+3(-5x)=6$

$⟹2x-15x=6$

$⟹x=\frac{-6}{13}$

$y=-5x=\frac{30}{13}$

$ForB:$

$y=\frac{x}{5}$

$⟹2x+\frac{3x}{5}=6$

$⟹x=\frac{5\times 3}{4}=\frac{15}{4}$

$y=\frac{x}{5}=\frac{3}{4}$

$\therefore A(\frac{-6}{13},\frac{30}{13})$

$B(\frac{15}{4},\frac{3}{4})$

$⟹OA=\sqrt{{\left(\frac{30}{13}\right)}^2+{\left(\frac{6}{13}\right)}^2}$

$OA=\sqrt{\frac{900+36}{{13}^2}}$

$OA=\frac{1}{13}\sqrt{936}$

$\because Isoscles,\therefore OA=OB=\frac{\sqrt{936}}{13}$

$Area=\frac{1}{2}\times OA\times OB=\frac{1}{2}\times \frac{936}{13\times 13}=\frac{36}{13}$

$A)Ans$