Parallel and Perpendicular Lines

Two lines that are parallel have the same slope.

Thus two lines given by $y=m_1x+b$y=m1​x+b and $y=m_2x+c$y=m2​x+c will be parallel if and only if $m_1=m_2$m1​=m2​.

Two lines are perpendicular if the product of their respective slopes equals $-1$−1. Another way to say this is that the slope of one line is the negative reciprocal of the other. 

Thus two lines given by $y=m_1x+b$y=m1​x+b and $y=m_2x+c$y=m2​x+c will be perpendicular if and only if $m_1=-\frac{1}{m_2}$m1​=−1m2​​.

We can explain these features with the following diagram:

The parallel property is self-evident. If the lines $PQ$PQ and $P'Q'$P′Q′ are parallel, then for a given run $q$q, the rise on both lines are equal. That is to say for both lines the slope $m=\frac{p}{q}$m=pq​. 

For the perpendicular property, let the positive slope of the line $BA$BA be $m_1=\frac{a}{b}$m1​=ab​. Now, imagine we rotate $BA$BA $90^\circ$90°  counterclockwise so that the new line $BA'$BA′ is perpendicular to $BA$BA. From the symmetry in the diagram we see that the slope of $BA'$BA′ is negative and given by $m_2=-\frac{b}{a}$m2​=−ba​. 

So for example two lines parallel to $y=2x+3$y=2x+3 are $y=2x+7$y=2x+7 and $y=2x-12$y=2x−12. 

Two lines perpendicular to $y=2x+3$y=2x+3 are $y=-\frac{1}{2}x+9$y=−12​x+9 and $y=-\frac{1}{2}x-1$y=−12​x−1.

There is a neat trick that allows you to immediately determine the slope of a line if its equation is in general form. Take for example the line with equation $2x-3y+7=0$2x−3y+7=0. Without actually doing it, if we were to isolate $y$y, we would add $3y$3y to both sides, and then divide both sides by $3$3. Therefore the coefficient of $x$x would become $\frac{2}{3}$23​, and this would be the lines slope. 

Take another example. The line given by $12x+3y-17=0$12x+3y−17=0 has a slope of $-4$−4, because, without actually doing it, we could subtract $12x$12x from both sides (note the $-17$−17 is irrelevant) and then divide by $3$3. Can you see that the slope of $5x-2y+9=0$5x−2y+9=0 is $\frac{5}{2}$52​? 

We might also mention a test for collinearity. Three points $A,B$A,B and $C$C are said to be collinear if they all lie on the same straight line. To test this, we could simply check that the slopes of $AB$AB and $AC$AC (or $AB$AB and $BC$BC for that matter) are equal. 

For example suppose we consider the three points $A\left(-3,7\right),B\left(1,12\right)$A(−3,7),B(1,12) and $\left(9,22\right)$(9,22).  The slope of the line segment $AB$AB is $m_1=\frac{12-7}{1-\left(-3\right)}=\frac{5}{4}$m1​=12−71−(−3)​=54​ and the slope of $AC$AC is $m_2=\frac{22-7}{9-\left(-3\right)}=\frac{15}{12}=\frac{5}{4}$m2​=22−79−(−3)​=1512​=54​. Hence the three points ae collinear.

Finally if we wish to find the vertical distance between two parallel lines, we simply find the distance between their y - intercepts. Note that this vertical distance is not normally the shortest distance between the two lines. To find the shortest or perpendicular distance between parallel lines we would need to make use of the perpendicular distance formula.  

Worked Examples

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