Question

The equation of the line perpendicular to the line $2x+3y-4=0$ and passing through the origin is -

Answer 1

$2x+3y-9=0$

$\Rightarrow 3y=-2x+4$

$\Rightarrow y=\frac{-2}{3}x+\frac{4}{3}$

Slope $=m,=\frac{-2}{3}$

Slope of the perpendicular to $m_1\Rightarrow m_1\times m_2=-1$

$m_2=\frac{1}{m_1}=\frac{3}{2}$

$m_2=3/2$

$y=m+c\Rightarrow y=\frac{3}{2}x+c$

it passes through original $\Rightarrow 0=\frac{3}{2}\left(0\right)+c\Rightarrow c=0$

$\therefore y=\frac{3}{2}x\Rightarrow 3x-2y=0$

$\therefore$ The required equation is $3x-2y=0$