Perpendicular lines — product of slopes = −1: Exercises

Before we start, one reminder of the two building blocks we lean on constantly:

Level 1 — Recognition

(Can you read off a slope and apply the test?)

Exercise 1.1

Line $A$ : $y = 5 x - 2$ . Line $B$ : $y = - \frac{1}{5} x + 9$ . Are they perpendicular?

Recall Solution 1.1

WHAT: read each slope, multiply, compare to $- 1$ .

From $y = m x + c$ : $m_{1} = 5$ , $m_{2} = - \frac{1}{5}$ .
Product: $5 \times (- \frac{1}{5}) = - 1$ . ✅

Answer: Yes, perpendicular.

Exercise 1.2

A line has slope $m = 7$ . What is the slope of any line perpendicular to it?

Recall Solution 1.2

WHAT: take the negative reciprocal.

Flip $7 \to \frac{1}{7}$ , then negate $\to - \frac{1}{7}$ .
Check: $7 \times (- \frac{1}{7}) = - 1$ . ✅

Answer: $m_{⊥} = - \frac{1}{7}$ .

Exercise 1.3

A line has slope $m = - \frac{3}{4}$ . Give the perpendicular slope.

Recall Solution 1.3

WHAT: flip the fraction, flip the sign.

Reciprocal of $- \frac{3}{4}$ is $- \frac{4}{3}$ ; negate it $\to + \frac{4}{3}$ .
Check: $- \frac{3}{4} \times \frac{4}{3} = - 1$ . ✅

Answer: $m_{⊥} = \frac{4}{3}$ .

Level 2 — Application

(Compute a slope first, then apply the test or build a line.)

Exercise 2.1

Line through $P (2, 1)$ and $Q (6, 4)$ . Find the slope of any line perpendicular to $P Q$ .

Recall Solution 2.1

WHAT: slope of $P Q$ first, then negative reciprocal.

$m_{1} = \frac{4 - 1}{6 - 2} = \frac{3}{4}$ . WHY: rise $\div$ run.
Perpendicular: $m_{⊥} = - \frac{1}{3/4} = - \frac{4}{3}$ .
Check: $\frac{3}{4} \times - \frac{4}{3} = - 1$ . ✅

Answer: $m_{⊥} = - \frac{4}{3}$ .

Exercise 2.2

Find the equation of the line through $(3, - 2)$ perpendicular to $y = \frac{1}{2} x + 4$ .

Recall Solution 2.2

WHAT: get perpendicular slope, then use point–slope form.

Given $m_{1} = \frac{1}{2}$ ⟹ $m_{⊥} = - 2$ .
WHY point–slope? We have one point and a slope — that's exactly $y - y_{0} = m (x - x_{0})$ . See Equation of a Line — point-slope form.
$y - (- 2) = - 2 (x - 3)$
$y + 2 = - 2 x + 6$
$y = - 2 x + 4$ .
Check: $\frac{1}{2} \times (- 2) = - 1$ . ✅

Answer: $y = - 2 x + 4$ .

Exercise 2.3

Is the line through $A (1, 1)$ , $B (4, 7)$ perpendicular to the line through $C (0, 5)$ , $D (6, 2)$ ?

Recall Solution 2.3

WHAT: two slopes, then multiply.

$m_{A B} = \frac{7 - 1}{4 - 1} = \frac{6}{3} = 2$ .
$m_{C D} = \frac{2 - 5}{6 - 0} = \frac{- 3}{6} = - \frac{1}{2}$ .
Product: $2 \times (- \frac{1}{2}) = - 1$ . ✅

Answer: Yes, perpendicular.

Level 3 — Analysis

(Unknowns inside the slope — solve for them.)

Exercise 3.1

For what value of $k$ is the line through $(1, 2)$ and $(k, 6)$ perpendicular to $y = 2 x + 1$ ?

Recall Solution 3.1

WHAT: the two-point slope must equal the required perpendicular slope, then solve for $k$ .

Target slope: perpendicular to $m = 2$ is $m_{⊥} = - \frac{1}{2}$ .
Two-point slope: $\frac{6 - 2}{k - 1} = \frac{4}{k - 1}$ .
Set equal: $\frac{4}{k - 1} = - \frac{1}{2}$ .
Cross-multiply: $4 \cdot 2 = - 1 \cdot (k - 1) \Rightarrow 8 = - (k - 1) = - k + 1$ .
$- k = 7 \Rightarrow k = - 7$ .

Answer: $k = - 7$ .

Exercise 3.2

The line $a x + 3 y = 7$ is perpendicular to $y = \frac{3}{4} x - 1$ . Find $a$ .

Recall Solution 3.2

WHAT: put the first line into slope form to read its slope, then apply the test.

$a x + 3 y = 7 \Rightarrow 3 y = - a x + 7 \Rightarrow y = - \frac{a}{3} x + \frac{7}{3}$ . So $m_{1} = - \frac{a}{3}$ .
Given $m_{2} = \frac{3}{4}$ . Perpendicular means $m_{1} m_{2} = - 1$ : $- \frac{a}{3} \cdot \frac{3}{4} = - 1 \Rightarrow - \frac{a}{4} = - 1 \Rightarrow a = 4.$
Check: $m_{1} = - \frac{4}{3}$ , and $- \frac{4}{3} \times \frac{3}{4} = - 1$ . ✅

Answer: $a = 4$ .

Exercise 3.3

Points $A (0, 0)$ , $B (4, 2)$ , $C (2, c)$ form a right angle at $B$ . Find $c$ .

Recall Solution 3.3

WHAT: the right angle sits at $B$ , so the two arms $B A$ and $B C$ must be perpendicular.

$m_{B A} = \frac{0 - 2}{0 - 4} = \frac{- 2}{- 4} = \frac{1}{2}$ .
$m_{B C} = \frac{c - 2}{2 - 4} = \frac{c - 2}{- 2}$ .
Perpendicular at $B$ : $m_{B A} \cdot m_{B C} = - 1$ : $\frac{1}{2} \cdot \frac{c - 2}{- 2} = - 1 \Rightarrow \frac{c - 2}{- 4} = - 1 \Rightarrow c - 2 = 4 \Rightarrow c = 6.$

Answer: $c = 6$ .

Level 4 — Synthesis

(Combine perpendicularity with lengths, midpoints, and full constructions.)

Exercise 4.1

Find the equation of the perpendicular bisector of the segment joining $P (2, 1)$ and $Q (8, 5)$ .

Figure — Perpendicular lines — product of slopes = −1

Recall Solution 4.1

WHAT: a perpendicular bisector passes through the midpoint of $P Q$ and is perpendicular to $P Q$ .

Midpoint $M = (\frac{2 + 8}{2}, \frac{1 + 5}{2}) = (5, 3)$ . WHY midpoint: "bisector" cuts $P Q$ in half.
Slope of $P Q$ : $m_{P Q} = \frac{5 - 1}{8 - 2} = \frac{4}{6} = \frac{2}{3}$ .
Perpendicular slope: $m_{⊥} = - \frac{3}{2}$ .
Point–slope through $M (5, 3)$ : $y - 3 = - \frac{3}{2} (x - 5) \Rightarrow y = - \frac{3}{2} x + \frac{15}{2} + 3 = - \frac{3}{2} x + \frac{21}{2} .$
Check: $\frac{2}{3} \times (- \frac{3}{2}) = - 1$ . ✅ And $M$ : $- \frac{3}{2} (5) + \frac{21}{2} = - \frac{15}{2} + \frac{21}{2} = 3$ . ✅

Answer: $y = - \frac{3}{2} x + \frac{21}{2}$ .

Exercise 4.2

Triangle with vertices $A (1, 1)$ , $B (5, 3)$ , $C (3, k)$ is right-angled at $A$ . Find $k$ , then the area of the triangle.

Recall Solution 4.2

WHAT: right angle at $A$ ⟹ $A B ⊥ A C$ ; then use the perpendicular legs as base and height.

$m_{A B} = \frac{3 - 1}{5 - 1} = \frac{2}{4} = \frac{1}{2}$ .
$m_{A C} = \frac{k - 1}{3 - 1} = \frac{k - 1}{2}$ .
Perpendicular: $\frac{1}{2} \cdot \frac{k - 1}{2} = - 1 \Rightarrow \frac{k - 1}{4} = - 1 \Rightarrow k - 1 = - 4 \Rightarrow k = - 3$ .
Now $C = (3, - 3)$ . The legs $A B$ and $A C$ are perpendicular, so area $= \frac{1}{2} ∣ A B ∣∣ A C ∣$ .
$∣ A B ∣ = (5 - 1)^{2} + (3 - 1)^{2} = 16 + 4 = 20$ (via Pythagoras Theorem/Distance Formula).
$∣ A C ∣ = (3 - 1)^{2} + (- 3 - 1)^{2} = 4 + 16 = 20$ .
Area $= \frac{1}{2} 20 \cdot 20 = \frac{1}{2} \cdot 20 = 10$ .

Answer: $k = - 3$ , area $= 10$ square units.

Level 5 — Mastery

(Multi-step reasoning; watch the degenerate cases.)

Exercise 5.1

Find the foot of the perpendicular from the point $P (4, 5)$ to the line $y = \frac{1}{2} x + 1$ .

Recall Solution 5.1

WHAT: drop a perpendicular from $P$ to the line $L$ ; the "foot" is where they meet. Plan: (1) build the perpendicular line through $P$ , (2) intersect it with $L$ .

$L$ has slope $m_{1} = \frac{1}{2}$ ⟹ perpendicular slope $m_{⊥} = - 2$ .
Perpendicular line through $P (4, 5)$ : $y - 5 = - 2 (x - 4) \Rightarrow y = - 2 x + 13$ .
Intersect with $L$ : set $\frac{1}{2} x + 1 = - 2 x + 13$ . $\Rightarrow \frac{1}{2} x + 2 x = 12 \Rightarrow \frac{5}{2} x = 12 \Rightarrow x = \frac{24}{5} = 4.8$ . $\Rightarrow y = - 2 (4.8) + 13 = - 9.6 + 13 = 3.4 = \frac{17}{5}$ .
Check the foot lies on $L$ : $\frac{1}{2} (4.8) + 1 = 2.4 + 1 = 3.4$ . ✅

Answer: Foot $= (\frac{24}{5}, \frac{17}{5}) = (4.8, 3.4)$ .

Exercise 5.2

Line $L_{1}$ is horizontal: $y = 3$ . Line $L_{2}$ passes through $(2, - 1)$ and is perpendicular to $L_{1}$ . Write the equation of $L_{2}$ . Why can't you use $m_{1} m_{2} = - 1$ here?

Recall Solution 5.2

WHAT: handle the degenerate horizontal/vertical case geometrically, not with the product rule.

$L_{1} : y = 3$ is horizontal, slope $m_{1} = 0$ .
A line perpendicular to a horizontal line is vertical — and a vertical line has undefined slope. You cannot write " $0 \times m_{2} = - 1$ ", because there is no finite $m_{2}$ : dividing $- 1$ by $0$ is impossible.
A vertical line through $(2, - 1)$ is just $x = 2$ (all points share $x = 2$ ).

Answer: $L_{2} : x = 2$ . The product rule fails because a vertical slope is undefined, so it must be treated as a special case — see the companion and Angle Between Two Lines for the $90°$ view.

Exercise 5.3

Show that the diagonals of the square with vertices $A (0, 0)$ , $B (4, 0)$ , $C (4, 4)$ , $D (0, 4)$ are perpendicular, using slopes.

Recall Solution 5.3

WHAT: the diagonals are $A C$ and $B D$ ; test their slopes.

$m_{A C} = \frac{4 - 0}{4 - 0} = 1$ .
$m_{B D} = \frac{4 - 0}{0 - 4} = \frac{4}{- 4} = - 1$ .
Product: $1 \times (- 1) = - 1$ . ✅

Answer: Perpendicular, since $m_{A C} m_{B D} = - 1$ .

Connections

Perpendicular Lines — Product of Slopes = −1 — the parent rule these exercises train.
Slope of a Line — every problem starts by reading or computing a slope.
Equation of a Line — point-slope form — used in 2.2, 4.1, 5.1.
Distance Formula & Pythagoras Theorem — lengths and the right-triangle area in 4.2.
Angle Between Two Lines — perpendicular is the $θ = 90°$ special case (relevant to 5.2).
Rotation of Vectors by 90° — the "why negative reciprocal" behind every answer.
Parallel Lines — equal slopes — the twin test $m_{1} = m_{2}$ .