Lesson 3 of 4Analytic Geometry

Unit 2 · Lesson 3 of 4

Slopes, Perpendicular Bisectors and Distance to a Line

Parallel lines share a slope and perpendicular lines have negative reciprocal slopes. Use those facts with the midpoint and length formulas to write the equation of a perpendicular bisector, and to find the shortest distance from a point to a line.

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Key things to remember

1Slope: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ (undefined for a vertical line, where $x_{2} = x_{1}$ ).
2Non-vertical parallel lines have equal slopes. Perpendicular lines that are neither horizontal nor vertical have negative reciprocal slopes, whose product is $- 1$ .
3The perpendicular bisector of $A B$ passes through the midpoint of $A B$ with the negative reciprocal of the slope of $A B$ ; every point on it is equidistant from $A$ and $B$ .
4The distance from a point to a line is measured along the perpendicular: build the perpendicular, find the foot by solving a system, then use the length formula.
5To find $b$ in $y = m x + b$ , substitute a known point and the slope, then solve.

Two towns want to share a new fire station, and they agree it must be the same distance from both. On a map in kilometres, the towns are at $A (- 3, - 1)$ and $B (5, 3)$ . The midpoint $(1, 1)$ is one fair spot, but it is not the only one: there is a whole line of fair spots, and finding its equation is the first goal of this lesson. The second goal is a different kind of shortest path. A farm at $P (7, 0)$ wants a straight driveway to a highway that runs along the line $y = 2 x + 1$ . Which point of the highway is closest, and how long is the driveway? Both problems are solved with slopes, perpendicular lines and the formulas from Lesson 1.

Slope and the equation of a line

The slope $m$ of a line measures its steepness: the rise divided by the run between any two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ on it, $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ . A line that rises to the right has a positive slope, one that falls to the right has a negative slope, a horizontal line has slope $0$ , and a vertical line has an undefined slope (its run is $0$ , and you cannot divide by $0$ ). A non-vertical line can be written as $y = m x + b$ , where $b$ is the y-intercept. To find $b$ when you know the slope and one point, substitute the point and solve: for slope $3$ through $(2, 1)$ , $1 = 3 (2) + b$ , so $b = 1 - 6 = - 5$ and the line is $y = 3 x - 5$ .

Parallel and perpendicular slopes

Two different non-vertical lines are parallel exactly when they have the same slope: they rise at the same rate, so the gap between them never changes. Two lines are perpendicular when they meet at a right angle ( $9 0^{\circ}$ ). For lines that are neither horizontal nor vertical, perpendicular slopes are negative reciprocals: flip the fraction and change the sign. So a line of slope $\frac{2}{3}$ is perpendicular to a line of slope $- \frac{3}{2}$ , and the product of the two slopes is $\frac{2}{3} \times (- \frac{3}{2}) = - 1$ . Here is why. A step along the first line of $3$ right and $2$ up, turned a quarter-turn counterclockwise, becomes a step of $2$ left and $3$ up. The turned line therefore has slope $\frac{3}{- 2} = - \frac{3}{2}$ : the run and rise trade places (the reciprocal) and one of them changes sign (the negative).

Slope of a line	Slope of a parallel line	Slope of a perpendicular line
$2$	$2$	$- \frac{1}{2}$
$- \frac{3}{4}$	$- \frac{3}{4}$	$\frac{4}{3}$
$1$	$1$	$- 1$
$- \frac{1}{5}$	$- \frac{1}{5}$	$5$
$0$ (horizontal)	$0$	Undefined (vertical line)

Perpendicular slopes multiply to negative 1, except for a horizontal line paired with a vertical line.

Horizontal and vertical lines

A horizontal line such as

y = 2

and a vertical line such as

x = - 1

are perpendicular, but the product rule cannot show it, because the vertical line has no slope to multiply. Recognize this pair by its equations: one has the form

y = c

and the other

x = d

, where

c

and

d

are constants.

The perpendicular bisector of a segment

The perpendicular bisector (also called the right bisector) of a segment is the line that passes through the segment's midpoint and is perpendicular to it. It has a special property: every point on it is the same distance from both endpoints, and every point that is the same distance from both endpoints lies on it. To see why, pick any point $P$ on the bisector of $A B$ , with midpoint $M$ . The triangles $PM A$ and $PMB$ share the side $PM$ , have $M A = MB$ , and both have a right angle at $M$ . They are congruent (identical in size and shape), so $P A = PB$ . That is why the bisector of the segment joining the two towns is the line of fair spots for the fire station.

Find the midpoint $M$ of the segment.
Find the slope of the segment.
Take the negative reciprocal to get the slope of the bisector.
Substitute $M$ and the new slope into $y = m x + b$ and solve for $b$ .
Check: $M$ satisfies the equation, and a second point on the bisector is equally far from both endpoints.

Fig 1Left: segment AB from A(−3, −1) to B(5, 3), its midpoint M(1, 1), and its perpendicular bisector y = −2x + 3; the point P(−1, 5) on the bisector is the square root of 40 from both A and B. Right: the shortest path from Q(4, −1) to the line y = 0.5x + 2 runs along a perpendicular and meets the line at F(2, 3); a vertical path to the same line is longer.

For the towns, $M = (\frac{- 3 + 5}{2}, \frac{- 1 + 3}{2}) = (1, 1)$ and the slope of $A B$ is $\frac{3 - ( - 1 )}{5 - ( - 3 )} = \frac{4}{8} = \frac{1}{2}$ . The bisector's slope is the negative reciprocal, $- 2$ . Substituting $M (1, 1)$ : $1 = - 2 (1) + b$ , so $b = 3$ , and the bisector is $y = - 2 x + 3$ . Test it with $P (- 1, 5)$ , which is on the line because $- 2 (- 1) + 3 = 5$ . Then $P A^{2} = (- 1 + 3)^{2} + (5 + 1)^{2} = 4 + 36 = 40$ and $P B^{2} = (- 1 - 5)^{2} + (5 - 3)^{2} = 36 + 4 = 40,$ so $P A = PB = 40 = 210 \approx 6.3$ km. A fire station at $P$ would be fair to both towns.

VideoFind the midpoint, take the negative reciprocal of the slope, and write the equation of the perpendicular bisector.Watch on YouTube

Shortest distance from a point to a line

The distance from a point to a line means the shortest distance, and for a point that is not on the line, the shortest path meets the line at a right angle. Any other path from the point to the line is the hypotenuse of a right triangle whose one leg is the perpendicular path, and a hypotenuse is longer than either leg. In the figure, the perpendicular path from $Q (4, - 1)$ has length $20 \approx 4.5$ , while the vertical path to the same line has length $5$ . To find the distance, build the perpendicular, find where it meets the line (a linear system, like those in Unit 1), and measure. Worked Example 4 uses these steps to show that the farm's driveway is $35 \approx 6.7$ km long.

Find the slope of the given line (rearrange to $y = m x + b$ if needed).
Write the equation of the line through the point with the negative reciprocal slope.
Solve the two equations as a system to find the intersection point $F$ , the foot of the perpendicular.
Use the length formula to find the distance from the point to $F$ .

Shortcut for horizontal and vertical lines

The perpendicular to a horizontal line is vertical, so the distance from

(3, 7)

y = 2

is just

7 - 2 = 5

. Likewise, the distance from

(3, 7)

to the vertical line

x = - 1

3 - (- 1) = 4

VideoFind the shortest distance from a point to a line by building the perpendicular line, finding where the two lines meet, and measuring the length.Watch on YouTube

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Line 1 passes through

(- 2, 1)

and

(4, 5)

. Line 2 passes through

(0, 6)

and

(2, 3)

. Are the lines parallel, perpendicular or neither?

Example 2

Find the equation of the line through

(4, - 1)

that is perpendicular to

y = 2 x - 5

Example 3

Find the equation of the perpendicular bisector of the segment from

A (- 3, - 1)

B (5, 3)

Example 4

A farm is at

P (7, 0)

and a straight highway follows

y = 2 x + 1

(units in kilometres). Find the shortest distance from the farm to the highway, exactly and to one decimal place.

Where marks get lost

Taking only half of "negative reciprocal": the slope perpendicular to $2$ is $- \frac{1}{2}$ , not $- 2$ (negative only) and not $\frac{1}{2}$ (reciprocal only).
Using the segment's own slope through the midpoint. That line is the segment's line, not its bisector. For $A (- 3, - 1)$ and $B (5, 3)$ , $y = \frac{1}{2} x + \frac{1}{2}$ is wrong; $y = - 2 x + 3$ is right.
Drawing the perpendicular through an endpoint instead of the midpoint. That line is perpendicular to the segment but does not bisect it.
Measuring the distance to a line along a vertical or horizontal path. From $P (7, 0)$ straight up to $y = 2 x + 1$ is $15$ km; the true (perpendicular) distance is $35 \approx 6.7$ km.
Stopping at the foot of the perpendicular. The intersection point $(1, 3)$ is where the shortest path ends; the distance is the length from $P$ to that point.

Flashcards 8

Slope formula

Parallel lines (not vertical)

Perpendicular lines (neither horizontal nor vertical)

Perpendicular bisector

+ 4 more

Practice

What is the slope of a line perpendicular to a line with slope

\frac{3}{4}

Find the slope of the line through

(- 2, 5)

and

(4, - 7)

Which pair of lines is perpendicular?

The line through

(6, 1)

perpendicular to

y = 3 x - 2

y = m x + b

. What is

b

Which is the equation of the perpendicular bisector of the segment from

A (2, 1)

B (6, 9)

Which point is the same distance from

A (- 3, - 1)

and

B (5, 3)

What is the distance from the point

(3, 7)

to the line

y = 2

units

Find the distance from

P (7, 2)

to the line

x + 3 y = 3

, to one decimal place.

units

The line through

(1, 2)

and

(4, k)

is perpendicular to a line with slope

\frac{3}{2}

. Find

k

Match each slope with the slope of a perpendicular line.

A. $- \frac{1}{5}$
B. $- 1$
C. $\frac{3}{2}$
D.Undefined (a vertical line)

5

- \frac{2}{3}

1

0

Put the steps for finding the distance from a point to a line in order.

1Use the length formula from the point to the foot
2Solve the system of the two lines to find the foot of the perpendicular
3Write the equation of the perpendicular line through the point
4Find the slope of the given line
5Take the negative reciprocal to get the perpendicular slope

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