Parallel and perpendicular lines and collinear points

Slopes make it easy to answer geo­met­ric ques­tions with a lit­tle arith­metic. Are two lines par­al­lel? Do they meet at right angles? Do three points lie on one line? Is a quadri­lat­eral a par­al­lel­o­gram, a rec­tan­gle or a rhom­bus? This les­son gives the slope con­di­tions for par­al­lel and per­pen­dic­u­lar lines and uses them, together with the dis­tance for­mula, to answer all these ques­tions.

Par­al­lel lines#

Two non-ver­ti­cal lines are par­al­lel exactly when they have the same incli­na­tion, and hence the same slope:

$$\ell_1 \parallel \ell_2 \iff m_1 = m_2$$

Two ver­ti­cal lines are also par­al­lel to each other, even though nei­ther has a slope.

Per­pen­dic­u­lar lines#

Two non-ver­ti­cal lines are per­pen­dic­u­lar exactly when the prod­uct of their slopes is $-1$:

$$\displaystyle \ell_1 \perp \ell_2 \iff m_1 m_2 = -1, \quad \text{that is,} \quad m_2 = -\frac{1}{m_1}$$

Why. If $\ell_1$ has incli­na­tion $\theta$ (acute, say), then a line per­pen­dic­u­lar to it has incli­na­tion $90° + \theta$. Its slope is $\displaystyle \tan(90° + \theta) = -\cot\theta = -\frac{1}{\tan\theta} = -\frac{1}{m_1}$.

In words: the slope of a per­pen­dic­u­lar line is the neg­a­tive rec­i­p­ro­cal. For exam­ple, a line of slope $\displaystyle \frac{2}{3}$ is per­pen­dic­u­lar to any line of slope $\displaystyle -\frac{3}{2}$.

A hor­i­zon­tal line (slope $0$) and a ver­ti­cal line (no slope) are also per­pen­dic­u­lar; this case is not cov­ered by the for­mula.

Collinear points#

Three points $A$, $B$, $C$ are collinear (lie on one straight line) exactly when the slope of $AB$ equals the slope of $BC$. The two seg­ments then have the same steep­ness and share the point $B$, so they lie on the same line.

Exam­ple 1#

Show that $A(1, 1)$, $B(3, 5)$ and $C(4, 7)$ are collinear.

$$\displaystyle \text{slope of } AB = \frac{5 - 1}{3 - 1} = 2, \qquad \text{slope of } BC = \frac{7 - 5}{4 - 3} = 2$$

The slopes are equal and $B$ is com­mon, so the points are collinear.

Exam­ple 2#

Show that $A(1, 2)$, $B(4, 5)$ and $C(6, 3)$ are the ver­tices of a right-angled tri­an­gle.

$$\displaystyle \text{slope of } AB = \frac{5 - 2}{4 - 1} = 1, \qquad \text{slope of } BC = \frac{3 - 5}{6 - 4} = -1$$

Their prod­uct is $1 \times (-1) = -1$, so $AB \perp BC$. The tri­an­gle has a right angle at $B$.

Exam­ple 3#

The line through $(2, k)$ and $(5, 7)$ is (a) par­al­lel (b) per­pen­dic­u­lar to the line through $(1, 3)$ and $(4, 9)$. Find $k$ in each case.

The sec­ond line has slope $\displaystyle \frac{9 - 3}{4 - 1} = 2$. The first has slope $\displaystyle \frac{7 - k}{5 - 2} = \frac{7 - k}{3}$.

(a) Par­al­lel: $\displaystyle \frac{7 - k}{3} = 2$, so $7 - k = 6$ and $k = 1$.

(b) Per­pen­dic­u­lar: $\displaystyle \frac{7 - k}{3} = -\frac{1}{2}$, so $\displaystyle 7 - k = -\frac{3}{2}$ and $\displaystyle k = \frac{17}{2}$.

Test­ing quadri­lat­er­als#

Slopes show which sides are par­al­lel or per­pen­dic­u­lar; the dis­tance for­mula shows which sides are equal. Recall that a par­al­lel­o­gram has both pairs of oppo­site sides par­al­lel, a rec­tan­gle is a par­al­lel­o­gram with a right angle, and a rhom­bus is a par­al­lel­o­gram with all sides equal (equiv­a­lently, with per­pen­dic­u­lar diag­o­nals).

Exam­ple 4#

Show that $A(0, 0)$, $B(4, 1)$, $C(6, 4)$, $D(2, 3)$, taken in order, form a par­al­lel­o­gram.

$$\displaystyle \text{slope } AB = \frac{1}{4}, \quad \text{slope } DC = \frac{4 - 3}{6 - 2} = \frac{1}{4}, \quad \text{slope } AD = \frac{3}{2}, \quad \text{slope } BC = \frac{4 - 1}{6 - 4} = \frac{3}{2}$$

Both pairs of oppo­site sides are par­al­lel, so $ABCD$ is a par­al­lel­o­gram. It is not a rec­tan­gle, because $\displaystyle \frac{1}{4} \times \frac{3}{2} \neq -1$, so adja­cent sides are not per­pen­dic­u­lar.

Exam­ple 5#

Show that the line join­ing $(2, -3)$ and $(-5, 1)$ is par­al­lel to the line join­ing $(7, -1)$ and $(0, 3)$.

The slopes are $\displaystyle \frac{1 - (-3)}{-5 - 2} = -\frac{4}{7}$ and $\displaystyle \frac{3 - (-1)}{0 - 7} = -\frac{4}{7}$. They are equal, so the lines are par­al­lel (they are not the same line, because the slope from $(2, -3)$ to $(7, -1)$ is $\displaystyle \frac{2}{5}$, which is dif­fer­ent).

Prac­tice#

1. Are lines with slopes $\displaystyle \frac{2}{3}$ and $\displaystyle -\frac{3}{2}$ per­pen­dic­u­lar?
2. Is the line through $(1, 2)$ and $(3, 8)$ par­al­lel to the line through $(0, -1)$ and $(2, 5)$?
3. Show that $(-2, -1)$, $(0, 3)$ and $(2, 7)$ are collinear.
4. Find $k$ so that $(1, 2)$, $(3, k)$ and $(5, 10)$ are collinear.
5. Show that $A(2, 1)$, $B(5, 4)$, $C(7, 2)$ form a right-angled tri­an­gle, and say where the right angle is.
6. The line through $(3, k)$ and $(6, 5)$ is per­pen­dic­u­lar to a line of slope $3$. Find $k$.
7. The line through $(k, 1)$ and $(2, 4)$ is par­al­lel to the line through $(0, 0)$ and $(3, 6)$. Find $k$.
8. Show that $A(1, 0)$, $B(4, 1)$, $C(5, 4)$, $D(2, 3)$ form a par­al­lel­o­gram, and that its diag­o­nals are per­pen­dic­u­lar. What kind of par­al­lel­o­gram is it?

Answers#