• # question_answer A pair of straight lines drawn through the origin form with the line $2x+3y=6$an isosceles right angled triangle, then the lines and the area of the triangle thus formed is                                                    [Roorkee 1993] A)            $x-5y=0$$5x+y=0$$\Delta =\frac{36}{13}$                            B)            $3x-y=0$$x+3y=0$$\Delta =\frac{12}{17}$ C)            $5x-y=0$$x+5y=0$$\Delta =\frac{13}{5}$                              D)            None of these

$y=mx$. It makes an angle of $\pm {{45}^{o}}$ with $2x+3y=6$.                    \ $\tan (\pm {{45}^{o}})=\frac{m-(-2/3)}{1+m(-2/3)}=\pm 1$                    or $3m+2=\pm (3-2m)$$\Rightarrow m=\frac{1}{5},-5$ Hence sides are                    $x-5y=0,$                    $5x+y=0$                    and $2x+3y=6$.                    Solving in pairs, vertices are $(0,\,0)$,                    $\left( \frac{6}{13},\frac{30}{13} \right)\,,\left( \frac{30}{13},-\frac{6}{13} \right)$.                    $\Delta =\left| \frac{1}{2}({{x}_{1}}{{y}_{2}}-{{x}_{2}}{{y}_{1}}) \right|=\frac{1}{2}\times \frac{936}{169}=\frac{36}{13}$.