Parallel and Perpendicular Lines: Free Response

The equation of $L$ is in standard form, so the coefficient of $x$ is not the slope. Solve for $y$ first:

$10y = -4x + 25, \qquad y = -\tfrac{2}{5}x + \tfrac{5}{2}.$

So $L$ has slope $-\tfrac{2}{5}$. A parallel line climbs at the same rate, so it has that same slope, $-\tfrac{2}{5}$.

A perpendicular line has the negative reciprocal. Flip $\tfrac{2}{5}$ to $\tfrac{5}{2}$ and change the sign, and since the slope was already negative, changing its sign makes the result positive:

$m_{\perp} = -\frac{1}{-\tfrac{2}{5}} = \tfrac{5}{2}.$

The product test confirms it:

$-\tfrac{2}{5} \cdot \tfrac{5}{2} = -\tfrac{10}{10} = -1.$

Get every slope into view before comparing anything.

Pair 1. Solve the first equation for $y$:

$-2y = -5x + 8, \qquad y = \tfrac{5}{2}x - 4.$

The two slopes are $\tfrac{5}{2}$ and $-\tfrac{5}{2}$. They are not equal, so the lines are not parallel. Multiply them:

$\tfrac{5}{2} \cdot \left(-\tfrac{5}{2}\right) = -\tfrac{25}{4},$

which is not $-1$, so they are not perpendicular either. Pair 1 is neither.

Pair 2. Solve the second equation for $y$:

$5y = -2x + 20, \qquad y = -\tfrac{2}{5}x + 4.$

The two slopes are $\tfrac{5}{2}$ and $-\tfrac{2}{5}$, which are not equal, so not parallel. Their product is

$\tfrac{5}{2} \cdot \left(-\tfrac{2}{5}\right) = -\tfrac{10}{10} = -1,$

so Pair 2 is perpendicular.

Write both operations out on a slope $m$. Changing the sign gives $-m$; the negative reciprocal gives $-\tfrac{1}{m}$. The product test asks for a product of exactly $-1$, which is two demands at once: the right sign, and the right size.

A sign change meets the first and ignores the second. On Pair 1's slope $\tfrac{5}{2}$ it produces $-\tfrac{5}{2}$, and

$\tfrac{5}{2} \cdot \left(-\tfrac{5}{2}\right) = -\tfrac{25}{4}.$

The sign is right and the size is $\tfrac{25}{4}$, not $1$. Reciprocating is the step that fixes the size, because a number times its reciprocal is $1$:

$\tfrac{5}{2} \cdot \left(-\tfrac{2}{5}\right) = -1.$

The two operations give the same answer exactly when $-m = -\tfrac{1}{m}$, that is when $m$ is its own reciprocal. Reciprocating changes the size of a number unless that size is already $1$, so this happens only for $m = 1$ and $m = -1$. (A slope of $0$ is not in the running at all: it has no reciprocal.) So a line of slope $1$ really is perpendicular to a line of slope $-1$, and there the shortcut is right by accident rather than by rule.

Each step is one negative reciprocal.

From $L$ to $M$, flip $\tfrac{3}{4}$ and change the sign:

$m_M = -\frac{1}{\tfrac{3}{4}} = -\tfrac{4}{3}.$

From $M$ to $N$, do the same to $-\tfrac{4}{3}$. Flipping gives $-\tfrac{3}{4}$, and changing the sign gives

$m_N = -\frac{1}{-\tfrac{4}{3}} = \tfrac{3}{4}.$

Each step passes the product test: $\tfrac{3}{4} \cdot \left(-\tfrac{4}{3}\right) = -1$ and $-\tfrac{4}{3} \cdot \tfrac{3}{4} = -1$. Two flips have carried the slope back to where it began, so $N$ has the same slope as $L$.

First, the slopes exist. A vertical line is perpendicular only to horizontal lines. So if $M$ were vertical, $L$ would have to be horizontal, with slope $0$, which the hypothesis rules out. Therefore $M$ is not vertical and has a slope, and the same argument covers $N$. Without this step the product test would be applied to a line that has no slope to supply.

Next, the equality. Call the two slopes $m_M$ and $m_N$. Perpendicularity to $L$ is the product test in each case:

$m \cdot m_M = -1, \qquad m \cdot m_N = -1.$

Since $m \neq 0$, each equation may be divided by $m$:

$m_M = -\frac{1}{m}, \qquad m_N = -\frac{1}{m}.$

The two right-hand sides are the same number, so $m_M = m_N$. Any two lines perpendicular to $L$ therefore have equal slopes. (Subtracting the two product equations gives it just as well: $m\left(m_M - m_N\right) = 0$ with $m \neq 0$ forces $m_M = m_N$.)

Last, what that settles. Equal slopes alone do not make two lines parallel. Writing $M$ as $y = m_M x + b_M$ and $N$ as $y = m_N x + b_N$, the slopes agree, so either the intercepts differ, in which case the lines never meet and are parallel, or the intercepts agree too and the two equations describe a single line. So the proved statement is this: $M$ and $N$ are parallel whenever they are two distinct lines.

Case 1: $L$ is horizontal, with slope $0$. The lines perpendicular to a horizontal line are the vertical lines $x = c$. Any two of those run straight up and down side by side, so they never meet unless $c$ is the same for both, in which case they are the same line. The conclusion of part B survives: two lines perpendicular to $L$ are parallel whenever they are distinct.

What settled it was inspection, not arithmetic. A vertical line has no slope, so there is no number to write in the second factor of

$m \cdot m_M = -1,$

and the equation cannot even be set up. The right angle between a horizontal line and a vertical line is read off the picture, exactly as it is for the two axes.

Case 2: $L$ is vertical, so $L$ itself has no slope. The lines perpendicular to a vertical line are the horizontal lines $y = k$, every one of them with slope $0$:

$m_M = 0 = m_N.$

Their slopes are equal, so the conclusion holds again, and two distinct horizontal lines are parallel. This time the perpendicular lines do have slopes, but the product test is still unavailable, because now it is $L$ that has no slope to contribute.

So the statement of part B is true more widely than its proof reaches. The proof needs $m$ to be a nonzero number; the two excluded cases have to be checked on their own, and both check out.

A street that never meets Elm Street runs in the same direction, so it has the same slope. Elm Street is in standard form, so solve for $y$ to see that slope:

$-4y = -3x + 12, \qquad y = \tfrac{3}{4}x - 3.$

The slope is $\tfrac{3}{4}$, and the new street carries that slope through the marker $(12, 5)$. Point-slope form:

$y - 5 = \tfrac{3}{4}(x - 12), \qquad y - 5 = \tfrac{3}{4}x - 9.$

Add $5$ to both sides:

$y = \tfrac{3}{4}x - 4.$

The two intercepts, $-3$ for Elm Street and $-4$ for the new street, are different, which is what makes this a second street rather than Elm Street written out again.

This street needs the negative reciprocal of Elm Street's slope $\tfrac{3}{4}$: flip it to $\tfrac{4}{3}$ and change the sign, giving $-\tfrac{4}{3}$. The product test checks it:

$\tfrac{3}{4} \cdot \left(-\tfrac{4}{3}\right) = -\tfrac{12}{12} = -1.$

Now point-slope form through $(12, 5)$:

$y - 5 = -\tfrac{4}{3}(x - 12), \qquad y - 5 = -\tfrac{4}{3}x + 16.$

Add $5$ to both sides:

$y = -\tfrac{4}{3}x + 21.$

A check on the marker: at $x = 12$ the right-hand side is $-16 + 21 = 5$, so the street does pass through it.

Two corners give the route's slope by the slope formula:

$m = \frac{1 - 9}{3 - (-3)} = \frac{-8}{6} = -\tfrac{4}{3}.$

That is the same slope as the street from part B, so the two are either parallel or the very same line; equal slopes on their own cannot tell those apart. Settle it by finding the route's own intercept, using the corner $(3, 1)$:

$1 = -\tfrac{4}{3}(3) + b = -4 + b, \qquad b = 5.$

So the route is $y = -\tfrac{4}{3}x + 5$, while the part B street is $y = -\tfrac{4}{3}x + 21$. Same slope, different intercepts: two different lines that never meet.

Read back into the town, that means the route and that street run alongside each other for their whole length. A rider who stays on the route never arrives at that street, however far they ride.

Test the lines in order, each against the one above it.

Line 1 is sound. Subtracting $2x$ from $2x + 3y = 9$ gives $3y = -2x + 9$, dividing by $3$ gives $y = -\tfrac{2}{3}x + 3$, and in that form the coefficient of $x$ is the slope:

$2x + 3y = 9 \;\Rightarrow\; y = -\tfrac{2}{3}x + 3.$

Line 2 is where the work breaks. It flipped $-\tfrac{2}{3}$ to $-\tfrac{3}{2}$ and never changed the sign. The product test exposes that at once:

$-\tfrac{2}{3} \cdot \left(-\tfrac{3}{2}\right) = \tfrac{6}{6} = 1,$

which is $1$, not $-1$. The negative reciprocal flips the fraction and changes the sign, and here the sign being changed is a minus, so line 2 should have read

$m_{\perp} = \tfrac{3}{2}, \qquad -\tfrac{2}{3} \cdot \tfrac{3}{2} = -1.$

Everything after line 2 is faithful to it. Line 3 substitutes $(-6, 4)$ into point-slope form with the correct sign inside the bracket, line 4 distributes $-\tfrac{3}{2}$ over $x + 6$ correctly, and line 5 adds $4$ to $-9$ correctly. A wrong number carried carefully forward is still wrong, which is exactly why the FIRST fault is the one worth finding.

With the slope corrected to $\tfrac{3}{2}$, redo the work from line 3 onwards. Point-slope form through $(-6, 4)$, where subtracting a negative coordinate leaves a plus sign inside the bracket:

$y - 4 = \tfrac{3}{2}\bigl(x - (-6)\bigr) = \tfrac{3}{2}(x + 6).$

Distribute, and note that $\tfrac{3}{2} \cdot 6 = 9$:

$y - 4 = \tfrac{3}{2}x + 9.$

Add $4$ to both sides:

$y = \tfrac{3}{2}x + 13.$

The perpendicularity check multiplies the two slopes:

$-\tfrac{2}{3} \cdot \tfrac{3}{2} = -\tfrac{6}{6} = -1,$

so the new line does cross the given one at a right angle.

The point check substitutes $x = -6$ into the answer:

$y = \tfrac{3}{2}(-6) + 13 = -9 + 13 = 4,$

which is the $y$-coordinate of the given point, so the line passes through $(-6, 4)$.

Now why the first check cannot stand alone. It looks only at the slope, and every line of the form

$y = \tfrac{3}{2}x + b$

has slope $\tfrac{3}{2}$ whatever $b$ is. That is a whole family of parallel lines, every one of them perpendicular to the given line, and every one of them passing the slope check. The given point is what selects a single member of that family: only one line of slope $\tfrac{3}{2}$ goes through $(-6, 4)$, and $b$ is exactly what the point determines. A check that every candidate passes cannot distinguish the right candidate from the rest.

Apply the slope formula to each pair of endpoints, taking the two points in the same order in the numerator and the denominator.

$m_{AB} = \frac{4 - 1}{4 - (-2)} = \frac{3}{6} = \tfrac{1}{2}, \qquad m_{BC} = \frac{8 - 4}{2 - 4} = \frac{4}{-2} = -2.$

$m_{CD} = \frac{5 - 8}{-4 - 2} = \frac{-3}{-6} = \tfrac{1}{2}, \qquad m_{DA} = \frac{1 - 5}{-2 - (-4)} = \frac{-4}{2} = -2.$

Only two numbers appear: $\tfrac{1}{2}$ on the sides $AB$ and $CD$, and $-2$ on the sides $BC$ and $DA$.

Three things have to be checked, and slopes can supply all three.

Opposite sides parallel. $AB$ and $CD$ both have slope $\tfrac{1}{2}$, and $BC$ and $DA$ both have slope $-2$.

Different lines, not one line. Equal slopes leave both possibilities open, so test an endpoint of one side against the other side's line. The line through $C(2, 8)$ with slope $\tfrac{1}{2}$ is

$y = \tfrac{1}{2}x + 7,$

and at $x = -2$ it gives $y = 6$, not the $1$ that belongs to $A$. So $A$ is off the line $CD$, and $AB$ and $CD$ are two different parallel lines. The same test on the other pair: the line through $A(-2, 1)$ with slope $-2$ is $y = -2x - 3$, and at $x = 4$ it gives $-11$, not the $4$ that belongs to $B$, so $BC$ and $DA$ are different lines too.

The corners. At every corner a side of slope $\tfrac{1}{2}$ meets a side of slope $-2$, and

$\tfrac{1}{2} \cdot (-2) = -1,$

so every corner is a right angle.

Both pairs of opposite sides are parallel, so $ABCD$ is a parallelogram, and its corners are right angles, so it is a rectangle. Notice what was never computed: no side length appears anywhere above, so this argument says nothing at all about whether the rectangle happens to be a square.

Look at what the four sides are. $PQ$ runs from $(2, 1)$ to $(7, 1)$: the $y$-coordinate never changes, so it is horizontal, with slope $0$. $QR$ runs from $(7, 1)$ to $(7, 5)$: the $x$-coordinate never changes, so it is vertical, and the slope formula would ask for

$m_{QR} = \frac{5 - 1}{7 - 7} = \frac{4}{0},$

which is not a number at all. $RS$ is horizontal like $PQ$, and $SP$ is vertical like $QR$.

Every corner of $PQRS$ joins a horizontal side to a vertical side, so at every corner one of the two slopes does not exist. The product test needs two numbers to multiply and there is no second number to hand it, so the test cannot be applied even once. Coming at it from the other side does not help either: the negative reciprocal of the horizontal side's slope $0$ would be $-\tfrac{1}{0}$, and dividing by zero is not allowed, so a slope of $0$ has no negative reciprocal to compare against.

What establishes the right angles is inspection. A horizontal line and a vertical line meet at a right angle, exactly as the two axes do. That is a fact about the picture rather than a consequence of any arithmetic on slopes.

So the rule has to be stated with its condition attached. For two lines that BOTH have slopes, the lines are perpendicular exactly when the product of the slopes is $-1$. A line with no slope is vertical, and it is perpendicular to exactly the lines of slope $0$, the horizontal ones. Stated that way the rule covers $ABCD$, where every side has a slope, and $PQRS$, where half of them do not.