# Parallel Lines Shortcut

Parallel lines in Analytical Geometry are two lines with the same slope. Basically, to solve for the equation of the line through a point and parallel to a given line with a known slope can be done in series of steps. However, using a shortcut formula, it can be done by one amazing move.

Derivation:

Given a line $ax+by+c=0$, find the equation of the line through  $(x_1,y_1)$.

The shortcut can be made if we observe a pattern when solving. Finding the line with a given conditions above has a pattern. Thus we can create a shortcut formula.

Step 1: Find the slope of the given line

$ax+by+c=0$

$by=-ax-c$

$y=\displaystyle\frac{-ax-c}{b}$

$y=\displaystyle\frac{-ax}{b}-\displaystyle\frac{c}{b}$

From here, the slope of the given line is  $\displaystyle\frac{-a}{b}$

Using the point-slope form, we can now solve for the equation of the required line.

$y-y_1=m(x-x_1)$

But  $m=\displaystyle\frac{-a}{b}$

$y-y_1=\displaystyle\frac{-a}{b}(x-x_1)$

$by-by_1=-ax+ax_1$

$ax+by=ax_1+by_1$

Worked Problem 1:

Find the equation of the line through  $(4,-2)$  and parallel to  $2x-y=4$.

Solution:

Using the derived formula above we have,

$ax+by=ax_1+by_1$

$(2)x+(-1)y=(2)(4)+(-1)(-2)$

$2x-y=10$

$2x-y-10=0$

Worked Problem 2:

Find the equation of the line through  $(5,4)$  and parallel to the line  $2x+9y=4$.

Solution:

Using the derived formula we have,

$ax+by=ax_1+by_1$

$(2)x+(9)y=(2)(5)+(9)(4)$

$2x+9y=46$

$2x+9y-46=0$

Worked Problem 3:

$ABCD$  is a rectangle with coordinates  $A(1,3),B(3,1),C(6,4)$.  Find the line containing the side  $\overline{AD}$.

Solution:

By drafting the figure as follows we have

We don’t have the coordinate of D but we can solve for the equation of the line $\overline{BC}$

Solving for slope of  $\overline{BC}$,

$m_{BC}=\displaystyle\frac{y_2-y_1}{x_2-x_1}$

$m_{BC}=\displaystyle\frac{4-1}{6-3}$

$m_{BC}=\displaystyle\frac{3}{3}$

$m_{BC}=1$

To solve for the equation of the line BC use either point B or C and use the point slope form.

$y-y_1=m(x-x_1)$

$y-1=1(x-3)$

$y-1=x-3$

$x-y-2=0$

We can now find the equation of the line AD using the shortcut formula above using point A and the line BC.