Lines, Slopes, and Perpendicularity

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Learning goals

Justify constant slope and the parallel criterion using similar right triangles
Prove the perpendicular slope criterion and handle horizontal and vertical cases
Find perpendicular bisectors from midpoint and direction information
Distinguish median and altitude equations using their geometric definitions
Locate intersections of geometric lines and check their defining conditions

A short line-equation refresher

From Algebra, the slope of a nonvertical line is its signed rise divided by its signed run. For distinct points

\begin{aligned}A&=(x_A,y_A),\\B&=(x_B,y_B)\end{aligned}

with different first coordinates,

m=\frac{y_B-y_A}{x_B-x_A}.

Keep the same subtraction order. A nonvertical line through $A$ has point-slope equation

y-y_A=m(x-x_A).

Solving for $y$ gives slope-intercept form. A vertical line instead fixes its first coordinate. Its slope is undefined, never infinity.

Worked example 1 Turn two points into a line equation

Problem. Find an equation for the line through

\begin{aligned}U&=(-5,7),\\V&=(3,3).\end{aligned}

Check both points.

Solution. The first coordinates differ, so the slope calculation is valid.

\begin{aligned} m&=\frac{3-7}{3-(-5)},\\ m&=-\frac12. \end{aligned}

Use $U$ in point-slope form, then clear the fraction.

\begin{aligned} y-7&=-\frac12(x+5),\\ x+2y&=9. \end{aligned}

At $U$ , the left side is $-5+2(7)$ , which is $9$ . At $V$ , it is $3+2(3)$ , also $9$ .

Why slope stays the same

Changing the size of a slope triangle changes both legs in the same proportion.

In the figure, the small triangle has rise $2$ and run $4$ . The larger one has rise $5$ and run $10$ . Both report slope $1/2$ . Their right angles agree, and their acute angles with the horizontal agree because their slanted sides lie on one line. AA similarity makes the matching legs proportional.

Matching acute angles and right angles make the two slope triangles similar.

Similarity makes slope independent of the chosen points#

Take two slope triangles on a falling line, with runs $8$ and $20$ . If the first signed rise is $-6$ , AA similarity multiplies its vertical leg by $20/8$ , giving signed rise $-15$ for the second. Both left-to-right trips fall, so

\begin{aligned} \frac{-6}{8}&=-\frac34,\\ \frac{-15}{20}&=-\frac34. \end{aligned}

For any nonhorizontal, nonvertical line, choose any two pairs of distinct points and travel left to right. Complete each pair to a right triangle with horizontal and vertical legs. The horizontal legs are parallel. The line makes equal corresponding acute angles with them, so AA makes the triangles similar. Hence the ratios of vertical leg length to horizontal leg length agree.

The positive runs give both signed rises the same sign: positive for a rising line and negative for a falling line. Thus the signed rise-to-run ratios also agree. Reversing either pair changes both signs and preserves its ratio. For a horizontal line, every rise is zero and every nonzero run gives slope zero. The particular leg lengths did not matter; the matching angles did.

Check your understanding

Two slope triangles on one falling line have runs 5 and 15. The smaller has signed rise $-2$ . What forces the larger rise to be $-6$ ?

A Their hypotenuses have equal lengths B AA similarity triples both leg lengths, and both rises are negative C Every right triangle has slope

-2

D Only the horizontal legs must triple

Parallel is a statement about distinct lines

Parallel nonvertical lines have the same slope. For example, slope triangles with runs $4$ and $6$ on two parallel lines have rises $2$ and $3$ . Corresponding angles make them similar, giving slope $1/2$ on both.

Corresponding angles on the parallel lines give similar slope triangles.

The parallel slope criterion and its converse#

Take distinct parallel rising lines. A slope triangle on the first has run $6$ and rise $5$ ; choose run $18$ on the second. Equal corresponding acute angles and right angles give AA similarity, so its rise is $15$ . Thus

\begin{aligned} \frac56&=\frac{15}{18}. \end{aligned}

For the forward direction in general, parallel slanted sides make equal angles with parallel horizontal legs. Together with the right angles, this gives AA similarity and equal rise-to-run ratios. This reasoning applies to any parallel slanted lines. If one parallel is horizontal, the other is horizontal too, so both slopes are zero.

For the converse, first compare two lines of slope $2/3$ , one through $(0,1)$ and the other through $(0,-2)$ . A shared point would require

\begin{aligned} \frac23x+1&=\frac23x-2,\\ 1&=-2. \end{aligned}

The impossible equality shows that they never meet. In general, two distinct nonvertical lines of the same slope have different intercepts. At a shared point, subtracting their equal slope terms would force the intercepts to agree. Hence they cannot meet and are parallel. The values of the intercepts were irrelevant; their difference mattered.

Thus two distinct nonvertical lines are parallel exactly when their slopes agree. Equal slopes and a shared point instead give one line, called coincident lines when described by two equations. Distinct vertical lines are parallel to each other; they have no slope numbers to compare.

Worked example 2 Check distinctness before concluding parallelism

Problem. Compare each line with $\ell$ , whose equation is

3x-2y=8.

Case A. $6x-4y=16$ .

Case B. $3x-2y=-2$ .

Solution. Case A divides by $2$ to give precisely the equation of $\ell$ . The equations describe the same line.

Case B has the same slope, $3/2$ , but a different intercept. Its intercept is $1$ , whereas $\ell$ has intercept $-4$ . These lines are distinct, so they are parallel.

Check your understanding

Two nonvertical lines both have slope $-3/4$ and contain $(2,5)$ . Which conclusion follows?

A They are distinct parallel lines B They are perpendicular C They coincide D Their relationship cannot be determined

A quarter-turn gives the perpendicular criterion

A right-angle turn exchanges the jobs of the horizontal and vertical legs, and one direction changes sign.

Start with run $3$ right and rise $2$ up. A $90^\circ$ counterclockwise rotation sends the rightward part upward and the upward part leftward. The rotated triangle has run $-2$ and rise $3$ , so its slope is $-3/2$ . The original slope is $2/3$ .

The quarter-turn sends a run of 3 and rise of 2 to a run of negative 2 and rise of 3.

Perpendicular slopes from rotated right triangles#

Use a fresh triangle with run $5$ right and signed rise $-8$ . A counterclockwise quarter-turn sends the rightward part upward and the downward part rightward. The new signed run is $8$ and rise is $5$ . The slopes multiply to

\left(-\frac85\right)\frac58=-1.

Now take a line that is neither horizontal nor vertical. Choose a positive run $r$ and a nonzero signed rise $s$ between two of its points. Its slope is

m=\frac{s}{r}.

Rotate the triangle $90^\circ$ counterclockwise about its starting point. The run becomes a rise of $r$ , and the signed rise becomes a run of $-s$ . This also covers a falling line: a downward step turns right, giving a positive run. The perpendicular direction has slope

\begin{aligned} n&=\frac{r}{-s},\\ mn&=-1. \end{aligned}

The rotation preserves the right triangle and turns its hypotenuse line through a right angle. The earlier uniqueness of a perpendicular through a point, and the perpendicular transversal theorem for parallels, show that any perpendicular line has this direction. Nothing depended on the chosen $r$ and $s$ .

Conversely, suppose two nonvertical, nonhorizontal lines have slopes whose product is $-1$ . Rotate the first through $90^\circ$ about one of its points. The image has the same slope as the second, so the image and second line coincide or are parallel. The original line is perpendicular to the image. In the parallel case the perpendicular transversal theorem transfers that right angle to the second line; in the coincident case it is already the same right angle. This proves the converse too.

For two nonvertical, nonhorizontal lines, perpendicularity is equivalent to slopes whose product is negative one. The slopes are negative reciprocals. Check both the sign and the reciprocal, rather than merely changing the sign.

Check your understanding

A slope triangle has run 7 and signed rise $-4$ . After a counterclockwise quarter-turn, what are its signed run and rise?

A Run

-7

, rise 4 B Run 4, rise 7 C Run

-4

, rise

-7

D Run 7, rise 4

Horizontal and vertical lines need their own case

A horizontal line and a vertical line meet at a right angle because their directions are parallel to the perpendicular coordinate axes. For example, the lines in the figure meet at $(-2,3)$ .

The right angle is valid even though one slope is undefined.

The horizontal slope is zero. The vertical slope is undefined, so there is no slope product to calculate. The negative reciprocal of zero also has no value. Through a specified point, the perpendicular to a horizontal line fixes the point’s first coordinate; the perpendicular to a vertical line fixes its second coordinate.

Perpendicular bisectors combine a point and a direction

The perpendicular bisector of a segment is the line through its midpoint perpendicular to its full line. For distinct endpoints, it passes through neither endpoint. From the earlier construction theorem, its points are exactly the points equidistant from the two endpoints.

Worked example 3 Find a perpendicular bisector

Problem. Find the perpendicular bisector of the segment with endpoints

\begin{aligned}A&=(-4,-1),\\B&=(2,3).\end{aligned}

Solution. Midpoint and slope give different pieces of information.

\begin{aligned} M&=(-1,1),\\ m_{AB}&=\frac23. \end{aligned}

The required line goes through $M$ with slope $-3/2$ .

\begin{aligned} y-1&=-\frac32(x+1),\\ 3x+2y&=-1. \end{aligned}

The solution line passes through M and is perpendicular to AB.

Substitution of $M$ gives $-1$ on the left of the final equation. The product of the two slopes is $-1$ , so the line meets both defining conditions.

Check your understanding

A segment has endpoints $(-6,4)$ and $(2,4)$ . Which equation is its perpendicular bisector?

y=4

x=-6

y=-2

x=-2

Medians, altitudes, and intersections

A median joins a triangle’s vertex to the opposite side’s midpoint. Its line needs those two points. An altitude passes through a vertex perpendicular to the opposite side’s full line. Its line needs the vertex and the perpendicular direction. The definitions do not normally select the same line.

Worked example 4 Separate a median from an altitude

Problem. Triangle $ABC$ has

\begin{aligned}A&=(0,6),\\B&=(-4,0),\\C&=(8,0).\end{aligned}

Find the median line and the altitude line from $A$ .

Solution. For the median, the midpoint of $BC$ is $M=(2,0)$ . The slope from $A$ to $M$ is $-3$ , giving

y=-3x+6.

For the altitude, $BC$ is horizontal. The perpendicular through $A$ is vertical, so its equation is

x=0.

The solid median reaches M; the dashed vertical altitude reaches a different point of BC.

The median passes through $M$ . The altitude meets $BC$ at $(0,0)$ , which is not its midpoint. A perpendicular line alone does not make a median.

An intersection must satisfy both line equations. Solve them together, then check the point in each. Parallel distinct lines have no intersection; coincident lines have infinitely many common points.

Worked example 5 Locate a median-altitude intersection

Problem. Triangle $ABC$ has

\begin{aligned}A&=(-2,5),\\B&=(0,-1),\\C&=(6,1).\end{aligned}

Find the intersection $P$ of the median line from $A$ and the altitude line from $B$ .

Solution. The midpoint of $BC$ is $M=(3,0)$ . The median through $A$ and $M$ has slope $-1$ :

y=-x+3.

The slope of $AC$ is $-1/2$ . Its perpendicular through $B$ has slope $2$ :

y=2x-1.

Set the heights equal and solve.

\begin{aligned} -x+3&=2x-1,\\ 3x&=4,\\ x&=\frac43,\\ y&=\frac53. \end{aligned}

The solid median and dashed altitude meet at P. The triangle sides identify which definition each line uses.

Thus $P=(4/3,5/3)$ . Substitution gives height $5/3$ in both equations. The median uses the correct midpoint, and the altitude slopes multiply to $-1$ , so the equations represent the requested geometric lines.

Check your understanding

A median line has equation $x=2$ .

Another vertex's altitude line has equation $y=-3x+10$ .

Where do the lines meet?

(2,4)

(4,2)

(2,10)

(0,10)

Lines, Slopes, and Perpendicularity

Learning goals

A short line-equation refresher

Why slope stays the same

Parallel is a statement about distinct lines

A quarter-turn gives the perpendicular criterion

Horizontal and vertical lines need their own case

Perpendicular bisectors combine a point and a direction

Medians, altitudes, and intersections

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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