Decide whether the given lines are parallel, perpendicular, or neither

The two lines in Figure 18 are parallel lines: they will never intersect. Notice that they have exactly the same steepness, which means their slopes are identical. The only difference between the two lines is the y-intercept. If we shifted one line vertically toward the y-intercept of the other, they would become the same line.

We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel.

Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is 1. So, if

m1 and m2{m}_{1}\text{ and }{m}_{2}m1​ and m2​

$-1$

m1m2=−1{m}_{1}{m}_{2}=-1m1​m2​=−1

To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is

18\frac{1}{8}81​

18\frac{1}{8}81​

As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes, assuming that the lines are neither horizontal nor perpendicular. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular.

{f(x)=14x+2negative reciprocal of14 is −4f(x)=−4x+3negative reciprocal of−4 is 14\begin{cases}f\left(x\right)=\frac{1}{4}x+2 & \text{negative reciprocal of}\frac{1}{4}\text{ is }-4 \\ f\left(x\right)=-4x+3 & \text{negative reciprocal of}-4\text{ is }\frac{1}{4} \end{cases}{f(x)=41​x+2f(x)=−4x+3​negative reciprocal of41​ is −4negative reciprocal of−4 is 41​​

The product of the slopes is –1.

−4(14)=−1-4\left(\frac{1}{4}\right)=-1−4(41​)=−1

Two lines are parallel lines if they do not intersect. The slopes of the lines are the same.

f(x)=m1x+b1 and g(x)=m2x+b2 are parallel if m1=m2f\left(x\right)={m}_{1}x+{b}_{1}\text{ and }g\left(x\right)={m}_{2}x+{b}_{2}\text{ are parallel if }{m}_{1}={m}_{2}f(x)=m1​x+b1​ and g(x)=m2​x+b2​ are parallel if m1​=m2​

If and only if

b1=b2{b}_{1}={b}_{2}b1​=b2​

m1=m2{m}_{1}={m}_{2}m1​=m2​

Two lines are perpendicular lines if they intersect at right angles.

f(x)=m1x+b1 and g(x)=m2x+b2 are perpendicular if m1m2=−1, and so m2=−1m1f\left(x\right)={m}_{1}x+{b}_{1}\text{ and }g\left(x\right)={m}_{2}x+{b}_{2}\text{ are perpendicular if }{m}_{1}{m}_{2}=-1,\text{ and so }{m}_{2}=-\frac{1}{{m}_{1}}f(x)=m1​x+b1​ and g(x)=m2​x+b2​ are perpendicular if m1​m2​=−1, and so m2​=−m1​1​

Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines.

{f(x)=2x+3h(x)=−2x+2g(x)=12x−4j(x)=2x−6\begin{cases}f\left(x\right)=2x+3 & & h\left(x\right)=-2x+2 \\ g\left(x\right)=\frac{1}{2}x - 4 & & j\left(x\right)=2x - 6 \end{cases}{f(x)=2x+3g(x)=21​x−4​​h(x)=−2x+2j(x)=2x−6​

Parallel lines have the same slope. Because the functions

$f\left(x\right)=2x+3$

$j\left(x\right)=2x-6$

12\frac{1}{2}21​

g(x)=12x−4g\left(x\right)=\frac{1}{2}x - 4g(x)=21​x−4

$h\left(x\right)=-2x+2$

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