# Prove that the line through the point (x1, y1)

Question:

Prove that the line through the point $\left(x_{1}, y_{1}\right)$ and parallel to the line $A x+B y+C=0$ is $A\left(x-x_{1}\right)+B\left(y-y_{1}\right)=0$.

Solution:

The slope of line $\mathrm{Ax}+\mathrm{B} y+\mathrm{C}=0$ or $y=\left(\frac{-\mathrm{A}}{\mathrm{B}}\right) x+\left(\frac{-\mathrm{C}}{\mathrm{B}}\right)$ is $m=-\frac{\mathrm{A}}{\mathrm{B}}$

It is known that parallel lines have the same slope.

$\therefore$ Slope of the other line $=m=-\frac{\mathrm{A}}{\mathrm{B}}$

The equation of the line passing through point $\left(x_{1}, y_{1}\right)$ and having a slope $m=-\frac{\mathrm{A}}{\mathrm{B}}$ is

$y-y_{1}=m\left(x-x_{1}\right)$

$y-y_{1}=-\frac{A}{B}\left(x-x_{1}\right)$

$\mathrm{B}\left(y-y_{1}\right)=-\mathrm{A}\left(x-x_{1}\right)$

$\mathrm{A}\left(x-x_{1}\right)+\mathrm{B}\left(y-y_{1}\right)=0$

Hence, the line through point $\left(x_{1}, y_{1}\right)$ and parallel to line $A x+B y+C=0$ is

$\mathrm{A}\left(x-x_{1}\right)+\mathrm{B}\left(y-y_{1}\right)=0$