Parallel lines — equal slopes: Exercises — Maths deep dive

Intuition How to use this page

These 12 problems climb a staircase: from L1 Recognition (spot equal slopes) to L5 Mastery (combine several ideas at once). Read the problem, cover the answer, forecast your steps, THEN open the collapsible solution. Every symbol used here was built in the parent note: slope $m = - a / b$ for $a x + b y + c = 0$ , and parallel $⟺ m_{1} = m_{2}$ (both lines non-vertical).

First, pin down one word we lean on constantly: the "intercept". When a line is written as $y = m x + c$ , the number $c$ is the == $y$ -intercept== — the height at which the line crosses the $y$ -axis (put $x = 0$ , you get $y = c$ ). Two lines with the same slope $m$ but different $c$ cross the $y$ -axis at different heights, so they run side by side forever: parallel-but-distinct. If they share BOTH $m$ and $c$ , they cross the $y$ -axis at the same point with the same tilt — they are literally the same line.

Keep this glowing the whole way down: two lines are ==parallel when they have the same slope $m$ but a different $y$ -intercept $c$ ==. The picture below is the mental model to carry through every level.

What to observe in the figure above: trace the cyan line ( $y = \frac{1}{2} x + 1$ ) and the amber line ( $y = \frac{1}{2} x + 4$ ). They rise at the identical steepness, so the white double-arrow between them measures a gap that never shrinks or grows — that constant gap is what "parallel" means. Notice the two lines cross the $y$ -axis at different heights ( $c = 1$ vs $c = 4$ ): same $m$ , different $c$ . Now watch the dashed white line ( $y = 1.4 x - 1$ ): it is steeper (bigger $m$ ), so it slices across the cyan line — different slope always means they eventually meet. Every problem on this page is about spotting or engineering that "same tilt, different intercept" condition.

Level 1 — Recognition

Goal: read a slope straight off a line, and check equal-slope by eye.

Recall Solution 1.1

What we do: turn every line into its slope $m = - a / b$ . Why? Because "parallel" only cares about direction, and direction $=$ slope.

$L$ : $a = 4, b = - 2 \Rightarrow m = - \frac{4}{- 2} = 2$ .
(a) $a = 2, b = - 1 \Rightarrow m = - \frac{2}{- 1} = 2$ . ✅ equal slope.
(b) $a = 2, b = 1 \Rightarrow m = - \frac{2}{1} = - 2$ . ❌ opposite tilt.
(c) $a = 8, b = - 2 \Rightarrow m = - \frac{8}{- 2} = 4$ . ❌ steeper. Answer: (a). It has slope $2$ but a different constant, so it is parallel, not identical.

Recall Solution 1.2

$L_{1}$ is already in slope–intercept form $y = m x + c$ , so $m_{1} = - \frac{3}{5}$ read directly.
$L_{2}$ : rearrange to $y = - \frac{3}{5} x + \frac{2}{5}$ , so $m_{2} = - \frac{3}{5}$ . (Or shortcut $m = - a / b = - \frac{3}{5}$ .) $m_{1} = m_{2} = - \frac{3}{5}$ ; their $y$ -intercepts differ ( $c = 8$ vs $c = \frac{2}{5}$ ) $\Rightarrow$ parallel. ✅

Level 2 — Application

Goal: build a specific parallel line, or solve for a parameter that forces parallelism.

Recall Solution 2.1

Step 1 — get the target slope. $m = - \frac{a}{b} = - \frac{5}{- 2} = \frac{5}{2}$ . Why? Parallel means we must copy this exact slope. Step 2 — point–slope form. Using $y - y_{1} = m (x - x_{1})$ from Point-Slope Form of a Line: $y - (- 3) = \frac{5}{2} (x - 2) ⟹ y + 3 = \frac{5}{2} (x - 2) .$ Step 3 — tidy. Multiply by $2$ : $2 y + 6 = 5 x - 10 \Rightarrow 5 x - 2 y - 16 = 0$ . Shortcut check: keep $5 x - 2 y = k$ , plug $(2, - 3)$ : $10 + 6 = 16$ , so $5 x - 2 y = 16$ . ✅ Same line.

Recall Solution 2.2

Slopes: $m_{1} = - \frac{k}{4}$ , and $m_{2} = - \frac{3}{- 6} = \frac{1}{2}$ . Parallel $⟺ m_{1} = m_{2}$ : $- \frac{k}{4} = \frac{1}{2}$ . Solve — and WHY each move is legal. Multiply both sides by $4$ (allowed: multiplying an equation by the same nonzero number keeps it true), giving $- k = 2$ . Multiply by $- 1$ (same reason): $k = - 2$ . Check it isn't the same line: with $k = - 2$ , $L_{1} : - 2 x + 4 y - 7 = 0$ , i.e. $x - 2 y + 3.5 = 0$ ; $L_{2} : x - 2 y + \frac{2}{3} = 0$ . Different constants $\Rightarrow$ genuinely parallel. ✅

Recall Solution 2.3

Step 1 — slope of the joining segment using $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ (the rise-over-run definition from Slope of a Line): $m = \frac{8 - 2}{4 - 1} = \frac{6}{3} = 2.$ Step 2 — copy the slope; parallel forces the same $m = 2$ . Why? the equal-slopes theorem. Step 3 — read the $y$ -intercept straight from the point. Our line has the form $y = m x + c = 2 x + c$ . It passes through $(0, 5)$ : since the point sits on the $y$ -axis ( $x = 0$ ), plugging in gives $5 = 2 (0) + c \Rightarrow c = 5$ . That is exactly the definition of the $y$ -intercept — the $y$ -value where $x = 0$ . Hence $y = 2 x + 5.$ ✅ (Equivalently via point–slope: $y - 5 = 2 (x - 0) \Rightarrow y = 2 x + 5$ — same answer.)

Level 3 — Analysis

Goal: reason about degenerate cases (vertical lines, hidden coincidence) and mixed conditions.

Recall Solution 3.1

$L_{2} : 2 x + 7 = 0 \Rightarrow x = - \frac{7}{2} = - 3.5$ . Both are vertical lines ( $x =$ constant). Slope trap: vertical lines have undefined slope — you cannot write $m_{1} = m_{2}$ as numbers. Handle by inspection instead. Since $- 3 \neq = - 3.5$ , they are two distinct vertical lines $\Rightarrow$ parallel (they never meet). If both constants were equal they'd be identical.

What to observe in the figure below: both the cyan line ( $x = - 3$ ) and the amber line ( $x = - 3.5$ ) shoot straight up — perfectly vertical, so no "rise per run" ratio exists (you'd divide by zero run). The white double-arrow shows a constant horizontal gap of $0.5$ . Because we cannot compare slopes as numbers, our only handle is the constants: $- 3 \neq = - 3.5$ , so distinct and parallel.

Recall Solution 3.2

Where the condition comes from. Parallel means equal slopes: $m_{1} = m_{2}$ . With $m = - a / b$ , $- \frac{2}{k} = - \frac{k + 1}{8} .$ Cancel the minus signs on both sides (multiply by $- 1$ ; keeps the equation true), leaving $\frac{2}{k} = \frac{k + 1}{8}$ . Before multiplying, secure the denominators. We may multiply both sides by $k$ and by $8$ only if neither is zero: $8 \neq = 0$ always, and we require $k \neq = 0$ (if $k = 0$ , $L_{1} : 2 x = 5$ is vertical with no slope — handled separately below). With $k \neq = 0$ confirmed, cross-multiply (one legal move that clears both fractions): $2 \cdot 8 = k (k + 1) ⟹ 16 = k^{2} + k ⟹ k^{2} + k - 16 = 0.$ This is exactly the coefficient rule $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}}$ from General Equation of a Line ax+by+c=0, now derived rather than quoted. $k = \frac{- 1 \pm 1 + 64}{2} = \frac{- 1 \pm 65}{2} .$ So $k = \frac{- 1 + 65}{2} \approx 3.531$ or $k = \frac{- 1 - 65}{2} \approx - 4.531$ .

Degenerate cases we set aside. $k = 0$ makes $L_{1}$ vertical (undefined slope) $\neq =$ $L_{2}$ 's finite slope, so no. $k = - 1$ makes $L_{2} : 8 y = 3$ horizontal (slope $0$ ), needing $m_{1} = - 2/ k = 0$ , impossible. Neither $0$ nor $- 1$ is a root, so nothing is lost. ✅

Coincidence check. Parallel needs $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ . Here $\frac{c _{1}}{c _{2}} = \frac{5}{3}$ , while $\frac{a _{1}}{a _{2}} = \frac{2}{k + 1}$ . For our $k$ values $\frac{2}{k + 1} \neq = \frac{5}{3}$ (e.g. $k \approx 3.531$ gives $\frac{2}{4.531} \approx 0.44 \neq = 1.67$ ), so both are true parallels, neither is coincident. ✅

Recall Solution 3.3

$A$ : set $y = 0 \Rightarrow 3 x = 12 \Rightarrow A = (4, 0)$ . $B$ : set $x = 0 \Rightarrow 4 y = 12 \Rightarrow B = (0, 3)$ . Slope of $A B = m = \frac{3 - 0}{0 - 4} = - \frac{3}{4}$ (matches $- a / b$ ). Through origin with same slope: $y = - \frac{3}{4} x$ , i.e. $3 x + 4 y = 0$ . Confirm not through $A, B$ : $A = (4, 0)$ : $3 (4) + 4 (0) = 12 \neq = 0$ . $B = (0, 3)$ : $0 + 12 = 12 \neq = 0$ . It passes through neither, as a parallel-but-distinct line must. ✅

Level 4 — Synthesis

Goal: fuse parallelism with distance, midpoints, and geometry of figures.

Recall Solution 4.1

A parallel line keeps $a, b$ : write it $12 x + 5 y + k = 0$ . Why? Same $a, b$ $\Rightarrow$ same slope $\Rightarrow$ parallel. Distance between parallels (from Distance Between Two Parallel Lines): $d = \frac{∣ c _{1} - c _{2} ∣}{a ^{2} + b ^{2}} = \frac{∣ - 3 - k ∣}{1 2 ^{2} + 5 ^{2}} = \frac{∣ - 3 - k ∣}{13} = 2.$ Why the absolute value? Distance is never negative; the constant we seek can sit on either side of the original line, so $∣ - 3 - k ∣$ captures both possibilities at once. So $∣ - 3 - k ∣ = 26 \Rightarrow - 3 - k = \pm 26 \Rightarrow k = - 29 or k = 23.$ Two answers (one on each side): $12 x + 5 y - 29 = 0$ and $12 x + 5 y + 23 = 0$ . ✅

Recall Solution 4.2

A parallelogram has opposite sides parallel. Compute the four side slopes with $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ :

$P Q$ : $\frac{3 - 1}{5 - 1} = \frac{2}{4} = \frac{1}{2}$ .
$S R$ : $\frac{6 - 4}{6 - 2} = \frac{2}{4} = \frac{1}{2}$ . → $P Q ∥ S R$ ✅
$QR$ : $\frac{6 - 3}{6 - 5} = \frac{3}{1} = 3$ .
$P S$ : $\frac{4 - 1}{2 - 1} = \frac{3}{1} = 3$ . → $QR ∥ P S$ ✅ Both pairs of opposite sides have equal slopes $\Rightarrow$ $P QR S$ is a parallelogram. ✅

What to observe in the figure below: the two cyan sides ( $P Q$ and $S R$ ) tilt at the identical gentle slope $\frac{1}{2}$ — you can see them running parallel across the shape. The two amber sides ( $QR$ and $P S$ ) share the steeper slope $3$ and likewise run parallel. Two matched parallel pairs is precisely the definition of a parallelogram, so the four labelled white vertices close into one.

Recall Solution 4.3

Midpoint $= (\frac{2 + 6}{2}, \frac{3 + ( - 5 )}{2}) = (4, - 1)$ . Keep $4 x - 3 y = k$ ; plug $(4, - 1)$ : $16 + 3 = 19$ . So $L_{2} : 4 x - 3 y = 19$ , i.e. $4 x - 3 y - 19 = 0$ . ✅

Level 5 — Mastery

Goal: multi-step problems where parallelism is one gear in a bigger machine.

Recall Solution 5.1

Step 1 — parallel form. Keep $a, b$ : $3 x + 4 y = c$ with $c > 0$ (to sit in the first quadrant). Step 2 — find axis intercepts. $x$ -intercept ( $y = 0$ ): $x = \frac{c}{3}$ . $y$ -intercept ( $x = 0$ ): $y = \frac{c}{4}$ . Step 3 — area of the right triangle formed with the axes $= \frac{1}{2} \cdot base \cdot height$ : $\frac{1}{2} \cdot \frac{c}{3} \cdot \frac{c}{4} = \frac{c ^{2}}{24} = 24 ⟹ c^{2} = 576 ⟹ c = 24 (c > 0) .$ Answer: $3 x + 4 y = 24$ . ✅

Recall Solution 5.2

Line 1 through origin, slope $\frac{3}{4}$ : $y = \frac{3}{4} x$ , i.e. $3 x - 4 y = 0$ . Line 2 parallel: $3 x - 4 y = k$ . Distance from line 1: $d = \frac{∣0 - k ∣}{3 ^{2} + ( - 4 ) ^{2}} = \frac{∣ k ∣}{5} = 5 ⟹ ∣ k ∣ = 25.$ Positive $y$ -intercept: line 2 is $y = \frac{3}{4} x - \frac{k}{4}$ ; intercept $- \frac{k}{4} > 0 \Rightarrow k < 0 \Rightarrow k = - 25$ . Line 2: $3 x - 4 y = - 25$ , i.e. $y = \frac{3}{4} x + \frac{25}{4}$ . $y$ -intercept $= \frac{25}{4} = 6.25$ . ✅

Recall Solution 5.3

Line through $A$ parallel to $2 x - y + 4 = 0$ . Its slope $m_{A} = - \frac{2}{- 1} = 2$ . Keep $2 x - y = k$ , plug $A (1, 2)$ : $2 (1) - 2 = 0$ . So $ℓ_{A} : 2 x - y = 0$ , i.e. $y = 2 x$ . Line through $B$ parallel to $x + 2 y - 5 = 0$ . Its slope $m_{B} = - \frac{1}{2}$ . Keep $x + 2 y = k$ , plug $B (3, 1)$ : $3 + 2 (1) = 5$ . So $ℓ_{B} : x + 2 y = 5$ . Intersection $P$ — solve the two together. Substitute $y = 2 x$ (from $ℓ_{A}$ ) into $ℓ_{B}$ : $x + 2 (2 x) = 5 ⟹ 5 x = 5 ⟹ x = 1, y = 2 (1) = 2.$ So $P = (1, 2)$ . (It coincides with $A$ , because $A$ already lies on $ℓ_{B}$ : $1 + 2 (2) = 5$ ✓.) Perpendicular check. Multiply the two constructed slopes: $m_{A} \cdot m_{B} = 2 \cdot (- \frac{1}{2}) = - 1.$ Product $= - 1$ $\Rightarrow$ the lines are perpendicular (the test from Perpendicular Lines — Product of Slopes = −1; see also Angle Between Two Lines). ✅

Active Recall

Recall Rapid re-derive (cover, forecast, reveal)

Slope of $a x + b y + c = 0$ ::: $m = - a / b$ (with $b$ carrying its own sign) Parallel form of $a x + b y + c = 0$ ::: keep $a, b$ , change constant: $a x + b y + k = 0$ Distance between $a x + b y + c_{1} = 0$ and $a x + b y + c_{2} = 0$ ::: $\frac{∣ c _{1} - c _{2} ∣}{a ^{2} + b ^{2}}$ Two vertical lines $x = p$ , $x = q$ with $p \neq = q$ are parallel — but WHY not via $m_{1} = m_{2}$ ? ::: because vertical slopes are undefined (no number); we judge by comparing the constants $p \neq = q$ , never by equating slopes Parallelogram test by slopes ::: both pairs of opposite sides have equal slopes

Connections

Parallel Lines — Equal Slopes
Slope of a Line
Point-Slope Form of a Line
General Equation of a Line ax+by+c=0
Distance Between Two Parallel Lines
Perpendicular Lines — Product of Slopes = −1
Angle Between Two Lines

Exercises — Parallel lines — equal slopes