Algebra 1 · Geometry · Grades 9, 10

Parallel and Perpendicular Lines: The Slope Rules

Quick answer

Two lines are parallel when their slopes are equal, and perpendicular when their slopes are negative reciprocals, meaning the two slopes multiply to -1. A line with slope 2 is parallel to any other line of slope 2, and perpendicular to any line of slope -1/2.

What you'll learn

  • Decide whether two lines are parallel, perpendicular or neither
  • Explain why perpendicular slopes are negative reciprocals
  • Write the equation of a line parallel or perpendicular to a given line

The two rules

RelationshipSlopesExample
Parallelequal: m1=m2m_1 = m_222 and 22
Perpendicularnegative reciprocals: m1m2=−1m_1 m_2 = -122 and −12-\tfrac{1}{2}

A negative reciprocal means flip the fraction and change the sign. Whole numbers count as fractions over 11, so the negative reciprocal of 3=313 = \tfrac{3}{1} is −13-\tfrac{1}{3}.

The reliable test is multiplication: if the product of the two slopes is −1-1, the lines are perpendicular.

Why parallel means equal slopes

Slope is the single number describing a line’s direction. Two lines that never meet must point the same way, and pointing the same way is exactly what equal slopes says.

Put the other way round: if two lines had different slopes, one would be climbing faster than the other, so the gap between them would keep changing — and a changing gap eventually reaches zero. Different slopes force an intersection. Equal slopes forbid one.

Why perpendicular means negative reciprocal

This rule looks arbitrary until you rotate a triangle.

Take a line of slope ab\tfrac{a}{b}: from any point, go across bb and up aa to reach another point on it. Draw that as a right triangle with legs bb and aa.

Now rotate the whole triangle a quarter turn. The leg that pointed across now points up, and the leg that pointed up now points across — but in the opposite direction. What was “across bb, up aa” becomes “across −a-a, up bb”.

So the rotated line has slope

riserun=b−a=−ba\frac{\text{rise}}{\text{run}} = \frac{b}{-a} = -\frac{b}{a}

The fraction flipped and the sign changed. That is the negative reciprocal, and it fell out of turning the triangle rather than from a rule anyone invented.

Multiplying confirms it:

ab×(−ba)=−abab=−1\frac{a}{b} \times \left(-\frac{b}{a}\right) = -\frac{ab}{ab} = -1

Worked examples

Common mistakes

Practice problems

  1. Are y=5x−2y = 5x - 2 and y=5x+7y = 5x + 7 parallel, perpendicular, or neither?

    Hint

    Compare the slopes, then the intercepts.

    Answer

    Parallel.

    Full solution

    Both slopes are 55 and the intercepts differ, so the lines are parallel and never meet.

  2. What is the perpendicular slope to m=37m = \tfrac{3}{7}?

    Answer

    −73-\tfrac{7}{3}

    Full solution

    Flip and negate: −73-\tfrac{7}{3}.

    Check: 37×−73=−1\tfrac{3}{7} \times -\tfrac{7}{3} = -1 ✓

  3. Are y=−4x+1y = -4x + 1 and y=14x−3y = \tfrac{1}{4}x - 3 perpendicular?

    Answer

    Yes.

    Full solution

    −4×14=−1-4 \times \tfrac{1}{4} = -1, so the lines are perpendicular.

  4. Are 3x+y=83x + y = 8 and y=3x+2y = 3x + 2 parallel?

    Hint

    Rearrange the first one before comparing.

    Answer

    No — they are neither parallel nor perpendicular.

    Full solution

    The first rearranges to y=−3x+8y = -3x + 8, so its slope is −3-3 while the second has slope 33.

    They are not equal, so not parallel. Their product is −9-9, not −1-1, so not perpendicular either.

  5. Write the equation of the line through (2,5)(2, 5) parallel to y=−2x+1y = -2x + 1.

    Answer

    y=−2x+9y = -2x + 9

    Full solution

    Same slope, −2-2. Point-slope: y−5=−2(x−2)=−2x+4y - 5 = -2(x - 2) = -2x + 4, so y=−2x+9y = -2x + 9.

    Check at x=2x = 2: y=5y = 5 ✓

  6. Write the equation of the line through (0,4)(0, 4) perpendicular to y=13x−6y = \tfrac{1}{3}x - 6.

    Answer

    y=−3x+4y = -3x + 4

    Full solution

    The negative reciprocal of 13\tfrac{1}{3} is −3-3. Since (0,4)(0, 4) is the intercept, y=−3x+4y = -3x + 4.

    Check: 13×−3=−1\tfrac{1}{3} \times -3 = -1 ✓

  7. Write the equation of the line through (−1,2)(-1, 2) perpendicular to y=25x+3y = \tfrac{2}{5}x + 3.

    Hint

    Find the negative reciprocal first, then use point-slope.

    Answer

    y=−52x−12y = -\tfrac{5}{2}x - \tfrac{1}{2}

    Full solution

    The perpendicular slope is −52-\tfrac{5}{2}.

    y−2=−52(x+1)=−52x−52y - 2 = -\tfrac{5}{2}(x + 1) = -\tfrac{5}{2}x - \tfrac{5}{2}

    So y=−52x−52+2=−52x−12y = -\tfrac{5}{2}x - \tfrac{5}{2} + 2 = -\tfrac{5}{2}x - \tfrac{1}{2}.

    Check at x=−1x = -1: 52−12=2\tfrac{5}{2} - \tfrac{1}{2} = 2 ✓

  8. A quadrilateral has vertices A(0,0)A(0,0), B(4,2)B(4,2), C(6,6)C(6,6) and D(2,4)D(2,4). Show that ABAB is parallel to DCDC.

    Hint

    Compute the slope of each side from its two endpoints.

    Answer

    Both have slope 12\tfrac{1}{2}, so they are parallel.

    Full solution

    Slope of ABAB: 2−04−0=12\tfrac{2 - 0}{4 - 0} = \tfrac{1}{2}.

    Slope of DCDC: 6−46−2=24=12\tfrac{6 - 4}{6 - 2} = \tfrac{2}{4} = \tfrac{1}{2}.

    The slopes are equal and the segments are not the same line, so AB∥DCAB \parallel DC.

    Checking the other pair, ADAD and BCBC both have slope 22, so this quadrilateral is a parallelogram.

Frequently asked questions

What exactly is a negative reciprocal?

Flip the fraction and change the sign. The negative reciprocal of 2, which is 2/1, is -1/2. The negative reciprocal of -3/4 is 4/3. A quick test: multiply the two slopes together and you should get -1.

Are two lines with the same slope always parallel?

Only if they are different lines. If the slope and the y-intercept both match, they are the same line lying on top of each other, which is usually counted separately from parallel.

What about horizontal and vertical lines?

They are perpendicular to each other, but the rule does not apply because a vertical line has no slope. You cannot multiply an undefined slope by anything, so treat that pair as a special case you recognise by sight.

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

  • CCSS.MATH.CONTENT.HSG.GPE.B.5Expressing Geometric Properties with EquationsProve the slope criteria for parallel and perpendicular lines and use them to solve geometric problems (e.g., find the equation of a line parallel or perpendicular to a given line that passes through a given point).
  • CCSS.MATH.CONTENT.HSA.CED.A.2Creating EquationsCreate equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.
For teachersTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.