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GuidePublished 15 Aug 20264 min readBy Kevin Joginparallel linesperpendicular linesslopenegative reciprocal
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KEVOS AIParallel and Perpendicular Lines

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Linear Functions

Parallel and Perpendicular Lines

Equal slopes for parallel, negative reciprocals for perpendicular, why the second rule holds, and the vertical-horizontal exception.

Category Engineering / MathematicsStream Linear FunctionsLevel CoreReading 5 minSource Week 5, pages 9-10

What this page covers

  • Test two lines for parallelism from their equations
  • Test two lines for perpendicularity
  • Find a line through a given point parallel or perpendicular to another
  • Handle the vertical and horizontal exception
On this page
  1. Parallel lines
  2. Perpendicular lines
  3. Why the product is -1
  4. The exception
  5. Worked example: a perpendicular through a point
  6. Common mistakes
  7. Frequently asked questions

Parallel lines

Parallel

Two distinct lines are parallel exactly when their slopes are equal. The source states it plainly: parallel lines have the same slope, or gradient.

It follows from the definition of slope as a rate. Two lines climbing at the same rate never converge, and two climbing at different rates must eventually meet.

The source's four parallel lines
EquationSlopey-intercept
y = 2x - 32-3
y = 2x + 121
y = 2x20
y = 2x - 42-4

All four have slope 2 and differ only in where they cross the y-axis. The source draws them as a family of parallel lines, which is exactly what varying b while fixing m produces.

Watch out

Equal slopes and equal intercepts means the same line, not two parallel ones. Lines are usually called parallel only when distinct.

Worked example — the source's case

Find the line through (-3, 5) parallel to y = 5x - 4.

The slope must be 5. Using point-slope form:

y - 5 = 5(x - (-3))y - 5 = 5(x + 3)y - 5 = 5x + 15y = 5x + 20Source example, Week 5, page 9
-5x + y = 20In standard form
Check

At x = -3: y = -15 + 20 = 5 ✓. The slope 5 matches the original, and the intercept 20 differs from -4, so the lines are distinct.

Perpendicular lines

Perpendicular
m1 · m2 = -1equivalently m2 = -1m1Source, Week 5, page 10

The second slope is the negative reciprocal of the first: invert the fraction and reverse the sign.

The source illustrates with a slope of 23 and its perpendicular -32, drawing the rise and run swapping roles as the line turns through a right angle.

Negative reciprocals
SlopePerpendicular slopeProduct
23-32-1
3-13-1
-133-1
-212-1
1-1-1

The source's own example is a line of slope 3 against one of slope -13, checking 3 × -13 = -1.

Why the product is -1

Take a line of slope pq, so travelling q right raises it by p. Rotating the whole picture through 90° turns the horizontal step into a vertical one and the vertical step into a horizontal one, with one of them reversing direction.

The rotated line therefore travels -p horizontally while rising q, giving slope q-p = -qp. Multiplying:

pq × (-qp) = -1Rise and run swap; one sign reverses
Note

The same statement appears in vector language: two lines are perpendicular when their direction vectors have zero dot product. For directions (q, p) and (-p, q) the dot product is -qp + pq = 0, which is the same fact. See The Dot Product.

The exception

Vertical and horizontal

A vertical line and a horizontal line are perpendicular, but the vertical one has undefined slope, so no product can be formed. The rule m1m2 = -1 does not apply and is not needed — the perpendicularity is obvious.

Every case
Line 1Line 2Parallel?Perpendicular?
y = mx + b1y = mx + b2Yes, if b1 ≠ b2Only if m2 = -1, impossible
y = mx + by = -1mx + cNoYes
x = ax = cYes, if a ≠ cNo
y = ky = cYes, if k ≠ cNo
x = ay = kNoYes — but the slope rule does not apply

Worked example: a perpendicular through a point

Perpendicular to 2x + 3y = 12 through (1, 4)

  1. Find the original slope. From Ax + By = C the slope is -AB = -23.
  2. Take the negative reciprocal. m = 32.
  3. Use point-slope form with the given point.
  4. Convert to whatever form is wanted.
y - 4 = 32(x - 1)2y - 8 = 3x - 33x - 2y = -5In standard form
Check

At (1, 4): 3 - 8 = -5 ✓. Slopes: -23 × 32 = -1 ✓.

Combining this with the midpoint formula gives the perpendicular bisector of a segment: find the midpoint, find the segment's slope, take the negative reciprocal, and use point-slope form.

Common mistakes

Errors and checks
MistakeCorrectCheck
Reciprocal without the sign changeNegative reciprocalThe product must be -1
Sign change without the reciprocalBoth are needed23 → -23 gives product -49
Reading the slope of Ax + By = C as ABIt is -ABRearrange to check
Applying the rule to a vertical lineNo slope existsPerpendicularity is obvious there
Calling identical lines parallelParallel lines are distinctCompare intercepts too
Using the wrong point in point-slope formUse the point the new line must pass throughSubstitute it back

Frequently asked questions

Why do perpendicular slopes multiply to -1?

Rotating a line through 90° swaps rise and run and reverses one sign. A slope of 23 becomes -32, and the product is -1.

Does the rule ever fail?

Yes, for a vertical and a horizontal line. They are perpendicular, but one has undefined slope so no product can be formed. It is the only exception.

How do I find a perpendicular slope quickly?

Turn the fraction upside down and change its sign. From 34 to -43; from -2 = -21 to 12.

Do parallel lines have the same intercept?

No — if they did they would be the same line. Parallel lines have equal slopes and different intercepts. The source draws four lines of slope 2 with intercepts -3, +1, 0 and -4.

Related pages

  • Slope of a Straight Line
  • The Four Forms of a Straight Line
  • The Cartesian Plane, Distance and Midpoint
  • The Dot Product and the Angle Between Vectors

Source. Handwritten teaching notes, Week 5, pages 9-10.

This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.

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