Parallel and perpendicular lines

In a nutshell

Lines can be parallel or perpendicular to each other. You can find out by looking at the gradients of the lines. If two lines are parallel, they share the same gradient. Two perpendicular lines will have the product of their gradients equal to $$-1$$. If you have the gradient of one line $$(m)$$, the gradient of a perpendicular line is equal to $$-\dfrac{1}{m}$$.

Parallel lines

Two lines that are parallel will have the same gradient. This is apparent if you look at a graph with parallel lines. They share the same value of $$m$$.

Example 1

Prove that the lines $$l_1\space (y=3x+2)$$​ and $$l_2\space (6x-2y -6=0)$$​ are parallel.

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Rearrange equation to give $$y=mx+c$$ . See if $$m$$ of $$l_1$$ is equal to $$m$$ of $$l_2$$.

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​$$\begin{aligned}l_1 \space m&=3\\l_2 \space is \space 6x-2y-6&=0\\-2y&=-6x+6\\2y&=6x-6\\y&=3x-3\\l_2\space m&=3\\ \end{aligned}$$

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​$$\underline{l_1= l_2=3}$$, therefore the lines are parallel.

Perpendicular lines

For perpendicular lines, the slope of one line is the negative reciprocal of its perpendicular counterpart. Take the gradient of one line and find the negative reciprocal $$\Big (-\dfrac{1}{m}\Big )$$ to find the gradient of the perpendicular line. The product of the gradients of the two perpendicular lines will equal $$-1$$, $$m_1\times m_2=-1$$.

Example 2

Investigate if the lines $$y=4x +2$$ and $$2x+8y-12=0$$ are perpendicular.

Rearrange equation to give $$y=mx+c$$.

$$\begin{aligned}2x+8y-12&=0\\8y&=-2x+12\\y&=-\dfrac28x+\dfrac{12}{8}\\\\y&=-\dfrac14x+\dfrac32\\\\m&=-\dfrac14\\\end{aligned}$$​​

Multiply the gradients to get $$-1$$.

​$$m_1 = 4, m_2=-\dfrac14$$

$$4\times -\dfrac14=-1$$

$$\underline{m_1 \times m_2=-1}$$, therefore the lines are perpendicular.

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