Perpendicular lines — product of slopes = −1: Worked examples

Everything here rests on one fact from the parent:

The scenario matrix

Below is the full menu of situations. Every cell is covered by at least one worked example (its label in brackets).

#	Case class	What makes it special	Covered by
C1	$m_{1} > 0$ (positive slope)	perpendicular slope is negative	Ex 1
C2	$m_{1} < 0$ (negative slope)	perpendicular slope is positive	Ex 2
C3	$m_{1}$ a fraction	flip and negate a fraction cleanly	Ex 3
C4	$m_{1} = 0$ (horizontal)	reciprocal undefined → vertical line	Ex 4
C5	vertical line (slope undefined)	rule doesn't apply → horizontal	Ex 4
C6	slope from two points	build $m_{1}$ first, then perpendicular	Ex 5
C7	equation of a perpendicular line through a point	combine with Equation of a Line — point-slope form	Ex 6
C8	real-world word problem	translate a story into slopes	Ex 7
C9	exam twist: unknown parameter	solve for a letter using $m_{1} m_{2} = - 1$	Ex 8
C10	geometry check: is a triangle right-angled?	test one pair of sides	Ex 9
C11	limiting behaviour ( $m_{1} \to \infty$ and $m_{1} \to 0$ )	what the negative reciprocal tends to	Ex 10

The worked examples

C1 — positive slope

Worked example Example 1 — Perpendicular of a positive slope

A line has slope $m_{1} = 4$ . Find the slope of any line perpendicular to it, and describe its direction.

Forecast: will the answer be positive or negative? Bigger or smaller than $1$ in size?

Write the rule. $m_{2} = - \frac{1}{m _{1}}$ . Why this step? The parent proved the perpendicular slope is the negative reciprocal — there is no shortcut worth memorising instead.
Substitute. $m_{2} = - \frac{1}{4}$ . Why this step? Reciprocal of $4$ is $\frac{1}{4}$ ; the minus is forced so the product hits $- 1$ , not $+ 1$ .
Read the direction. $m_{1} = 4$ climbs steeply up (go $1$ right, up $4$ ). $m_{2} = - \frac{1}{4}$ drifts gently down (go $4$ right, down $1$ ). Why this step? Perpendicular means "turned a quarter-circle", which swaps steep-up for gentle-down — sanity we can see, not just compute.

Verify: $4 \times (- \frac{1}{4}) = - 1$ . ✅

C3 — a fraction, done carefully

C4 & C5 — the degenerate cases (horizontal ⟂ vertical)

Here the formula $m_{2} = - 1/ m_{1}$ cannot be used, so we reason from the picture instead.

Figure — Perpendicular lines — product of slopes = −1

Worked example Example 4 — Horizontal meets vertical

Line $A$ is $y = 5$ (a flat horizontal line). What line through the origin is perpendicular to it?

Forecast: try plugging $m_{1} = 0$ into $- 1/ m_{1}$ . What goes wrong?

Find $m_{1}$ . $y = 5$ never rises: rise $= 0$ for any run, so $m_{1} = 0$ . Why this step? We always start from the actual slope, and horizontal lines have zero rise.
Try the formula — and watch it break. $m_{2} = - \frac{1}{0}$ , which is undefined (you cannot divide by zero). Why this step? We must see the rule fail here so we know to switch methods, not force a wrong number.
Reason geometrically instead. Look at the figure: the flat line is crossed at $90°$ by a line pointing straight up — a vertical line, e.g. $x = 0$ . A vertical line has slope undefined (run $= 0$ , so rise/run divides by zero). Why this step? Two lines can be perpendicular even when one slope is undefined — the eye confirms what the algebra can't.
State the special rule. Horizontal ( $m = 0$ ) $⊥$ vertical (undefined). Always true, never via the product formula.

Verify: The $x$ -axis ( $y = 0$ ) and $y$ -axis ( $x = 0$ ) meet at the origin at a right angle — a fact you can see without any multiplication. ✅ (No product of slopes exists to check, which is exactly the point.)

C6 — slope from two points

C7 — full equation of a perpendicular line

Worked example Example 6 — Perpendicular line through a given point

Find the equation of the line through $(6, - 2)$ that is perpendicular to $y = \frac{3}{4} x + 9$ .

Forecast: the given slope is $\frac{3}{4}$ — jot down what the perpendicular slope must be before continuing.

Read the given slope. In $y = m x + c$ , the number multiplying $x$ is the slope, so $m_{1} = \frac{3}{4}$ . Why this step? The $y = m x + c$ form hands us the slope directly.
Perpendicular slope. $m_{2} = - \frac{1}{3/4} = - \frac{4}{3}$ . Why this step? Flip $\frac{3}{4} \to \frac{4}{3}$ , negate — negative reciprocal.
Point–slope form. Using $(x_{0}, y_{0}) = (6, - 2)$ : $y - (- 2) = - \frac{4}{3} (x - 6)$ . Why this step? Equation of a Line — point-slope form is fastest when we know one point and the slope.
Tidy up. $y + 2 = - \frac{4}{3} x + 8 ⟹ y = - \frac{4}{3} x + 6$ . Why this step? Expanding $- \frac{4}{3} \times (- 6) = 8$ , then move the $+ 2$ across.

Verify: slopes $\frac{3}{4} \times (- \frac{4}{3}) = - 1$ ✅. Point check: at $x = 6$ , $y = - \frac{4}{3} (6) + 6 = - 8 + 6 = - 2$ ✅ (matches the given point).

C8 — real-world word problem

Worked example Example 7 — Ship's rescue course

A coastguard boat sails on a straight course rising $3$ km north for every $6$ km east. A rescue vessel must set off from port and cross the coastguard's track at a perfect right angle to reach it fastest. What slope must the rescue vessel steer?

Forecast: the coastguard climbs gently; will the rescue vessel's course go up or down as it heads east?

Coastguard slope. $m_{1} = \frac{north}{east} = \frac{3}{6} = \frac{1}{2}$ . Why this step? "north per east" is exactly rise over run — a slope.
Right-angle crossing ⟹ perpendicular slope. $m_{2} = - \frac{1}{1/2} = - 2$ . Why this step? "Cross at a right angle" is the physical meaning of perpendicular, so the product-of-slopes rule applies.
Interpret. $m_{2} = - 2$ : for every $1$ km east, the rescue vessel goes $2$ km south. It cuts steeply across the gentle coastguard track. Why this step? Translating the number back into a real heading is the whole reason we did the maths.

Verify: $\frac{1}{2} \times (- 2) = - 1$ . ✅

C9 — exam twist: solve for an unknown

C10 — geometry: is the triangle right-angled?

Worked example Example 9 — A right angle hiding in a triangle

Triangle has vertices $A (1, 1)$ , $B (5, 3)$ , $C (3, 7)$ . Show that the angle at $B$ is a right angle.

Forecast: which two sides meet at $B$ ? Those are the ones to test.

Slope of side $B A$ . $m_{B A} = \frac{1 - 3}{1 - 5} = \frac{- 2}{- 4} = \frac{1}{2}$ . Why this step? The angle at $B$ is squeezed between sides $B A$ and $B C$ , so those are the two slopes we compare.
Slope of side $B C$ . $m_{B C} = \frac{7 - 3}{3 - 5} = \frac{4}{- 2} = - 2$ . Why this step? Same reason — the second arm of the angle at $B$ .
Test the product. $m_{B A} \cdot m_{B C} = \frac{1}{2} \times (- 2) = - 1$ . Why this step? Product $= - 1$ is the exact fingerprint of a right angle between the two arms.

Verify (independent, via Pythagoras Theorem): $∣ B A ∣^{2} = (1 - 5)^{2} + (1 - 3)^{2} = 16 + 4 = 20$ ; $∣ B C ∣^{2} = (3 - 5)^{2} + (7 - 3)^{2} = 4 + 16 = 20$ ; $∣ A C ∣^{2} = (3 - 1)^{2} + (7 - 1)^{2} = 4 + 36 = 40$ . Then $20 + 20 = 40$ ✅ — Pythagoras agrees, so the angle at $B$ is genuinely $90°$ .

C11 — limiting behaviour

Worked example Example 10 — What happens at the extremes?

Describe the perpendicular slope as (a) the original line gets steeper and steeper ( $m_{1} \to \infty$ ), and (b) the original line flattens out ( $m_{1} \to 0^{+}$ ).

Forecast: as a line becomes nearly vertical, its perpendicular becomes nearly…?

Steep case. As $m_{1} \to \infty$ , $m_{2} = - \frac{1}{m _{1}} \to 0$ . Why this step? Dividing $1$ by a huge number gives almost zero — and a nearly-vertical line is crossed at right angles by a nearly-horizontal one. The picture matches the limit.
Flat case. As $m_{1} \to 0^{+}$ (positive but tiny), $m_{2} = - \frac{1}{m _{1}} \to - \infty$ . Why this step? Dividing $1$ by a tiny positive number gives a huge value; the minus makes it plunge to $- \infty$ — a nearly-flat line is crossed by a nearly-vertical one. This is exactly the degenerate C4/C5 boundary being approached smoothly.

Verify (numeric sample): take $m_{1} = 1000$ : $m_{2} = - 0.001$ , tiny as predicted ✅. Take $m_{1} = 0.001$ : $m_{2} = - 1000$ , huge as predicted ✅. Both still satisfy $m_{1} m_{2} = - 1$ .

Active Recall

Recall Cover and answer

Perpendicular slope of $m_{1} = - \frac{2}{7}$ ?
Which two cases make the product rule unusable, and what do you do instead?
If $y = k x - 1 ⊥ y = \frac{2}{3} x + 4$ , what is $k$ ?
In Example 9, which two side-slopes did you multiply, and to what value?

Answers: 1) $+ \frac{7}{2}$ . 2) Horizontal ( $m = 0$ ) and vertical (undefined) — check geometrically, they're perpendicular. 3) $k = - \frac{3}{2}$ . 4) $m_{B A} = \frac{1}{2}$ and $m_{B C} = - 2$ , product $= - 1$ .

Connections

Perpendicular Lines — Product of Slopes = $-1$ — the parent this page drills.
Slope of a Line — every example starts by reading or building a slope.
Parallel Lines — equal slopes — the contrast case ( $m_{1} = m_{2}$ ).
Equation of a Line — point-slope form — powers Example 6.
Angle Between Two Lines — perpendicular is the $90°$ special case.
Rotation of Vectors by 90° — why "flip and negate" turns a direction a quarter-circle.
Distance Formula & Pythagoras Theorem — the independent check in Example 9.

Worked examples — Perpendicular lines — product of slopes = −1

The scenario matrix

The worked examples

C1 — positive slope

C2 — negative slope

C3 — a fraction, done carefully

C4 & C5 — the degenerate cases (horizontal ⟂ vertical)

C6 — slope from two points

C7 — full equation of a perpendicular line

C8 — real-world word problem

C9 — exam twist: solve for an unknown

C10 — geometry: is the triangle right-angled?

C11 — limiting behaviour

Active Recall

Connections