Question

A pair of perpendicular straight lines is drown through the origin forming with the line $2x+3y=6$ an isosceles triangle right angled at the origin. Find the equation of pair of lines and the area of triangle.

Answer 1

Let any line through origin be $y=mx$. It makes an angle
of $\pm {45}^o$ with $2x+3y=6$ whose slope is $-2/3$
$\therefore \quad \tan { \left( \pm 45^{ o } \right) =
\dfrac { m-\left( -\dfrac { 2 }{ 3 } \right) }{ 1+m\left(
-\dfrac { 2 }{ 3 } \right) } = } \dfrac { 3m+2 }{ 3-2m }$
$\therefore (3-2m{)}^2=(3m+2{)}^2$.
Now put $m=\frac{y}{x}$
$\therefore \quad \left( 3-2\dfrac { y }{ x } \right) ^{
2 }=\left( 3\dfrac { y }{ x } +2 \right) ^{ 2 }$
or $(3x-2y{)}^2=(3y+2x{)}^2$
or $5x^2-24xy-5y^2=0$
is the required equation of the pair of lines through
origin and making an angle of ${45}^o$ with the given
line.
Area of $△=\frac{1}{2}BC.p=BD.p.$
But $p=BD\tan {45}^o=BD$
$\therefore Area=p^2,p$ is perpendicular from
origin on the line $2x+3y=6$
$\therefore \quad p=\dfrac{ 6 }{ \surd(4+9) }=\dfrac{ 6 }{
\surd 13 }$.
$\therefore Area=\frac{36}{13}sq.$ units.