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Analytic Geometry: (lesson 3 of 4)

## Parallel and Perpendicular Lines

Given two lines

\begin{aligned} &l_1: \ \ y = m_1 \cdot x + c_1 \\ &l_2: \ \ y = m_2 \cdot x + c_2 \end{aligned}

If $l_1$is perpendicular to $l_2$, then: $m_1 = m_2$.

Example 1:

$y = 3x + 14$ is parallel to $y = 3x - 72$

If two lines are not parallel, then there exists a point of intersection. This point can be found by solving the two equations simultaneously.

Example2:

Determine whether the following pairs of lines are parallel.

$l_1$: $y = x + 6$

$l_2$: the line joining $A$ $(1, 4)$ and $B$ $(-4, -1)$

Solution:

Gradient of $l_1 = 1$

Gradient of $l_2 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 4}{-4 - 1} = 1$

Since the two gradients are the same, the lines are parallel.

### Perpendicular Lines

Given two lines

\begin{aligned} &l_1: \ \ y = m_1 \cdot x + c_1 \\ &l_2: \ \ y = m_2 \cdot x + c_2 \end{aligned}

If $l_1$ is perpendicular to $l_2$, then: $m_1 m_2 = -1$.

Example 3:

$y = 3x + 14$ is perpendicular to $y = - \frac{1}{3} - 72$

Example 4:

Given the line $2x - 3y = 9$ and the point $(4, -1)$, find lines through the point that are

1: parallel to the given line and

2: perpendicular to it.

Solution for parallel line:

Clearly, the first thing needed is to solve $2x - 3y = 9$ for $y =$, so that the reference slope can be found:

$2x - 3y = 9 - 3y = -2x + 9 y = \frac{2}{3} x - 3$

So the reference slope from the reference line is $m = \frac{2}{3}$.

Since a parallel line has an identical slope, then the parallel line through $(4, -1)$ will have slope $m = frac{2}{3}$. So now, a point and a slope are known! So the point-slope form can be, now, used to find the line:

$y - (-1) = \frac{2}{3}(x - 4)$

$y + 1 = \frac{2}{3}x - \frac{8}{3}$

$y = \frac{2}{3}x - \frac{8}{3} - \frac{3}{3}$

$y = \frac{2}{3}x - \frac{11}{3}$

This is the parallel line that was asked for.

Solution for perpendicular line:

For the perpendicular line, the perpendicular slope has to be found. The reference slope is $m = \frac{2}{3}$, and, for the perpendicular slope, this slope will be flipped and the sign changed. So, the perpendicular slope is $m = - \frac{3}{2}$. Now, the slope-intercept form could be used.

\begin{aligned} y - y_1 &= m (x - x_1) \\ y - (-1) &= - \frac{3}{2}(x - 4) \\ y + 1 &= - \frac{3}{2}x + 6 \\ y &= - \frac{3}{2}x + 5 \end{aligned}

Example 5:

Find the perpendicular bisector of the line segment joining $A$ $(-3, 4)$ and $B$ $(2, -1)$.

Solution:

Gradient of $AB = \frac{4 + 1}{-3 - 2} = - 1$

\begin{aligned} m_1m_2 &= -1 \\ -1 \cdot m_2 &= -1 \\ m_2 &= 1 \end{aligned}
Midpoint of $AB = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{-3 + 2}{2}, \frac{4 - 1}{2} \right) = \left( -\frac{1}{2}, \frac{3}{2} \right)$
\begin{aligned} y - y_1 &= m \cdot (x - x_1) \\ y - \frac{3}{2} &= 1 \cdot (x - \frac{1}{2}) \\ y &= x - \frac{1}{2} + \frac{3}{2} \\ y &= x + 1 \end{aligned}