Question

Find the equation of the straight line through the point $(\alpha ,\beta )$ and perpendicular to the line $lx+my+n=0.$

Answer 1

Any line is given by equation

$y-y_1=m(x-x_1)...\left(1\right)$

where $\left(x_1{y}_1\right)$ is $(\alpha ,\beta )$

And m is negative reciprocal of slope of

line $lm+my+n=0$

i.e., $y=\frac{-lx}{m}-\frac{n}{m}$

$\Rightarrow$ Slope of line $=\frac{-l}{m}$

$\Rightarrow$ Putting the data in (1), we get

$y-\beta =\frac{m}{l}(x-\alpha )$

$ly+mx=m\alpha +l\beta$

$m(x-\alpha )=l(y-\beta )$