Question

A pair of perpendicular straight lines through the origin form an isosceles triangle with line $2x+3y=6$, then area of the triangle so formed is

• A. $\frac{36}{13}$
• B. $\frac{12}{17}$
• C. $\frac{13}{5}$
• D. $\frac{17}{13}$

Answer 1

Answer: (A)

$\frac{36}{13}$

In the above question, the $2x+3y=6$ is the third side of the isosceles triabgle

Let us assume that the perpendicular lines be $y=mx$ and $y=\frac{-x}{m}$

Since they are mutually perpendicukar to one another and they pass through origin.Hence the angle between the lines $y=mx$ and $2x+3y=6$

will be ${45}^0$ or ${135}^0$

Since $\tan \left({45}^0\right)=1$ and $\tan \left({135}^0\right)=-1$ hence $m^{\prime }=\pm 1$

Therefore $m^{\prime }=\pm 1=\frac{\frac{2}{3}+m}{1-\frac{2m}{3}}$

Hence $m=-5$ or $m=-\frac{1}{5}$

Thus perpendicular line are $-5y=x$ and $y=5x$

From the point of intersection of all three lines we get the vertices of the triangle, which are $(0,0),(\frac{-6}{14},\frac{30}{14}),(\frac{30}{13},\frac{6}{13})$

Thus area would be $\frac{bh}{2}$

H is the perpendicular distance of line $2x+3y=6$ from origin $=\frac{6}{\sqrt{13}}$

And base is (by distance formula) is $=\frac{12}{\sqrt{13}}$

Hence area$=\frac{36}{13}$