# Parallel and Perpendicular Lines

## Parallel Lines

Parallel lines are two or more lines that are always the same distance apart and never meet, no matter how far they are extended. In other words, they have the same slope and will never intersect. For example, the tracks of a railway are parallel lines.

If the slope coefficients of two lines are known and equal to k1 and k2, then the parallelism condition for these lines has the form

$k_1 = k_2$

If two lines have the general equations

$a_1x + b_1y + c_1 = 0,\;\;a_2x + b_2y + c_2 = 0,$

then their slopes k1 and k2 are respectively

$k_1 = -\frac{a_1}{b_1},\;\;k_2 = -\frac{a_2}{b_2}.$

For these lines to be parallel, their slopes must be equal, which means that:

$-\frac{a_1}{b_1} = -\frac{a_2}{b_2}$

This can be rearranged to give:

${a_1}{b_2} = {b_1}{a_2}$

So, if the coefficients of x and y in the general equations of two lines satisfy the above condition, then the lines are parallel.

## Perpendicular Lines

Perpendicular lines are lines that intersect at a right angle (90 degrees). They have slopes that are negative reciprocals of each other: For example, a vertical line and a horizontal line intersect at right angles and are therefore perpendicular.

$k_1 = -\frac{1}{k_2}$

or equivalently

$k_1k_2 = -1$

For example, the sides of a square are perpendicular to each other.

Another example is the x-axis and the y-axis: The x-axis and the y-axis are two lines that intersect at a right angle (90 degrees) and divide the coordinate plane into four quadrants.

Two straight lines

$a_1x + b_1y + c_1 = 0 \;\;\text{and}\;\;a_2x + b_2y + c_2 = 0$

are perpendicular if

$a_1a_2 + b_1b_2 = 0$

## Solved Problems

### Example 1.

Show that the lines $${2x - 3y + 4 = 0}$$ and $${6x - 9y - 2 = 0}$$ are parallel.

Solution.

Convert the given equations into slope-intercept form:

$2x - 3y + 4 = 0,\;\Rightarrow 3y = 2x - 4, \;\Rightarrow y = \frac{2}{3}x - \frac{4}{3}.$
$6x - 9y - 2 = 0,\;\Rightarrow 9y = 6x - 2, \;\Rightarrow y = \frac{6}{9}x - \frac{2}{9} = \frac{2}{3}x - \frac{2}{9}.$

So from here we get

$k_1 = k_2 = \frac{2}{3}.$

Hence, the lines are parallel.

### Example 2.

Show that the lines $${3x - 5y + 1 = 0}$$ and $${10x + 6y - 7 = 0}$$ are perpendicular.

Solution.

Rewrite the equations of the lines in slope-intercept form:

$3x - 5y + 1 = 0,\;\Rightarrow 5y = 3x + 1, \;\Rightarrow y = \frac{3}{5}x + \frac{1}{5}.$
$10x + 6y - 7 = 0,\;\Rightarrow 6y = -10x + 7, \;\Rightarrow y = -\frac{10}{6}x + \frac{7}{6} = -\frac{5}{3}x + \frac{7}{6}.$

Here we have

$k_1 = \frac{3}{5},\;k_2 = -\frac{5}{3}.$

Since

$k_1k_2 = \frac{3}{5} \cdot \left({-\frac{5}{3}}\right) = -1,$

the straight lines are perpendicular.

### Example 3.

Find the equation of the line passing through the point of intersection of the lines $${2x - 3y - 1 = 0}$$ and $${3x - y - 2 = 0}$$ and perpendicular to the line $${x - y + 1 = 0}.$$

Solution.

To find the point of intersection of the lines, we solve the system of equations

$\begin{cases} 2x - 3y - 1 = 0 \\ 3x - y - 2 = 0 \end{cases}, \;\Rightarrow \begin{cases} y = \frac{2}{3}x - \frac{1}{3} \\ y = 3x - 2 \end{cases}.$

Determine the $$x-$$coordinate of the intersection point:

$\frac{2}{3}x - \frac{1}{3} = 3x - 2,\;\Rightarrow 2x - 1 = 9x - 6, \;\Rightarrow 7x = 5, \;\Rightarrow x = \frac{5}{7}.$

Then the $$y-$$coordinate is

$y = 3x - 2 = 3\cdot\frac{5}{7} - 2 = \frac{15}{7} - 2 = \frac{1}{7},$

that is, the intersection point has coordinates

$x_0 = \frac{5}{7},\;y_0 = \frac{1}{7}.$

Our line must be perpendicular to the line $${x - y + 1 = 0}$$ which has the slope $$k_1 = 1.$$ Determine the slope $$k_2$$ of our straight line from the condition of perpendicularity:

$k_2 = -\frac{1}{k_1} = -\frac{1}{1} = -1.$

Now we can easily write the equation of the line using the point-slope form:

$y - y_0 = k_2\left({x - x_0}\right).$

Therefore

$y - \frac{1}{7} = -1\left({x - \frac{5}{7}}\right),\;\Rightarrow y - \frac{1}{7} = -x + \frac{5}{7},\;\Rightarrow 7y - 1 = -7x + 5, \;\Rightarrow 7x + 7y - 6 = 0.$

### Example 4.

Find the equation of the line passing through the point $$A\left({-3,4}\right)$$ and parallel to the line ${\frac{x-1}{2} = \frac{y+2}{3}}.$

Solution.

Remember that if a line is defined in the point-direction form

${\frac{x - x_0}{X} = \frac{y - y_0}{Y}},$

then the vector $$\mathbf{s}\left(X, Y\right)$$ is directed along the straight line, and the point $$\left(x_0, y_0\right)$$ lies on the line.

It is clear that a line parallel to

${\frac{x - 1}{2} = \frac{y + 2}{3}}$

and passing through another point $$A\left(x_1, y_1\right) = A\left(-3, 4\right)$$ will have the equation

${\frac{x - x_1}{X} = \frac{y - y_1}{Y}}, \;\Rightarrow \frac{x + 3}{2} = \frac{y - 4}{3}.$

### Example 5.

Find the equation of the line passing through the point $$A\left({-3,4}\right)$$ and perpendicular to the line ${x = t + 3,\; y = -7t + 4.}$

Solution.

The straight line is given by a parametric equation. Let's calculate its slope:

$\begin{cases} x = t + 3 \\ y = -7t + 4 \end{cases},\;\Rightarrow \begin{cases} t = x - 3 \\ y = -7\left({x-3}\right) + 4 \end{cases},\;\Rightarrow y = -7x + 21 + 4, \;\Rightarrow y = -7x+25.$

So the slope is $$k = -7.$$ Using the point-slope form, we get

$y - y_0 = k\left({x - x_0}\right),\;\Rightarrow y - 4 = -7\left({x+3}\right),\;\Rightarrow y - 4 = -7x - 21 \;\text{ or }\;y = -7x - 17.$

### Example 6.

At what value of $$a$$ the lines $$ax - 4y = 6$$ and $$x - ay = 3$$ are parallel?

Solution.

Rearrange the given equations and write them in slope-intercept form:

$ax - 4y = 6,\;\Rightarrow 4y = ax - 6, \;\Rightarrow y = \frac{a}{4}x - \frac{3}{2};$
$x - ay = 3,\;\Rightarrow ay = x - 3, \;\Rightarrow y = \frac{1}{a}x - \frac{3}{a}.$

The straight lines are parallel when

$k_1 = k_2,\;\Rightarrow \frac{a}{4} = \frac{1}{a},\; \Rightarrow a^2 = 4, \; \Rightarrow a = \pm2.$

Check each value of $$a.$$ When $$a = 2:$$

$\begin{cases} 2x - 4y = 6 \\ x - 2y = 3 \end{cases},\;\Rightarrow \begin{cases} x - 2y = 3 \\ x - 2y = 3 \end{cases},$

that is, in this case, the two lines coincide.

Now substitute the value $$a = -2:$$

$\begin{cases} -2x - 4y = 6 \\ x + 2y = 3 \end{cases},\;\Rightarrow \begin{cases} x + 2y = -3 \\ x + 2y = 3 \end{cases}.$

Therefore, we have two different parallel lines at $$a = -2.$$