Quarry School

Write a parallel or perpendicular line through a point

Explain it like I am five

Picture choosing a road's direction and then moving it until it passes through your house. The direction alone does not pick one road; many separated roads can share it. The house address finishes the choice. For a parallel line, keep the given slope. For a perpendicular line, use its negative reciprocal. Then use the supplied point to find the starting height. Point-slope form is a way to write this immediately: measure every new point's across and upward changes from the supplied point. Those changes must follow the chosen slope. You can leave the answer in that form or expand it to y = mx + b.

123456−4−224681012(0, −19)(3.17, 0)(4, 5)
The new line passes through the marked point; compare its slope with the dashed given line.
Reminder
  • Slope from two points. From (−2, 6) to (4, 5), run = 4 − (−2) = 6 and rise = 5 − 6 = −1.
  • Negative reciprocal. −16 becomes 6; their product is −1.
  • Point-slope form. Through (4, 5) at slope 6: y − 5 = 6(x − 4).
  • Distribution. 6(x − 4) = 6x − 24 because 6 multiplies both terms.
  • Solving for the intercept. 5 = 24 + b gives b = −19; plugging back gives 5 = 24 − 19.
  • Multiplying signed numbers. (−13) × (−3) = 1, so distributing the negative slope can produce a positive constant.
Why it works. If the line must pass through (x1, y1), its slope equation is m = y−y1x−x1 for other points with a different input. Multiplying gives y − y1 = m(x − x1). This equation also includes the supplied point because both changes are 0 there. It is point-slope form. Alternatively substitute that point in y = mx + b and solve for b. Both methods force the chosen slope and the required point to belong to one line.
RuleRule: Choose m equal to the given slope for a parallel line, or m = −1mgiven for a perpendicular line with nonzero finite given slope. Through (x1, y1), use y − y1 = m(x − x1).
The same idea, five ways
Say it

Say: choose the direction, then anchor it at the supplied address.

Write it

One point and one slope determine one nonvertical line.

In math
  • y − y1 = m(x − x1)
  • b = y1 − mx1
  • parallel: m = mgiven
  • perpendicular: m = −1mgiven
Like

Slide a road with the correct direction until it passes through your house.

See it
−22−224681012(0, 4)(−1.33, 0)required point
The new line passes through the marked point; compare its slope with the dashed given line.
The same idea, other ways
From changes

Through (1, 7) at slope 3, every new rise y − 7 equals three times the new run x − 1. That gives y − 7 = 3(x − 1).

y − 7 = 3(x − 1)
rise from the point = slope × run from the point
The supplied point becomes the starting address for measuring changes.
From the starting height

Write y = 3x + b and insert (1, 7): 7 = 3 + b, so b = 4. Plug back in: 3 × 1 + 4 = 7.

224681012(0, 4)(−1.33, 0)found startrequired point
The new line passes through the marked point; compare its slope with the dashed given line.
.1Parallel through a point

Keep the old line's slope, then adjust the starting height until the new line reaches the required address. If that address is already on the old line, this construction gives the old line itself; a distinct parallel line cannot pass through a point of the old line.

  • Rule: A parallel construction uses the same slope.
  • Rule: b = y1 − mx1 selects the intercept that fits the point.
  • Rule: Different intercepts are required for distinct parallel lines.
−22−224681012(0, 4)(−1.33, 0)(1, 7)
The new line passes through the marked point; compare its slope with the dashed given line.
Reminder
  • Distribution. 3(x − 1) = 3x − 3.
  • Substitution. The point (1, 7) means use x = 1 and y = 7 together.
The same idea, five ways
Say it

Say: keep the direction and find the starting height.

Write it

A parallel line through a point outside the given line has the same slope and a different intercept.

In math
  • m = mgiven
  • y − y1 = m(x − x1)
Like

Move a straight rail sideways without turning it.

See it
−22−224681012(0, 4)(−1.33, 0)(1, 7)
The new line passes through the marked point; compare its slope with the dashed given line.
Worked exampleThe verified parallel line

You need a line with the same direction as f(x) = 3x + 6 that passes through (1, 7).

−22−224681012(0, 4)(−1.33, 0)(1, 7)
The new line passes through the marked point; compare its slope with the dashed given line.
  1. Keep m = 3. Write y − 7 = 3(x − 1).Parallel direction uses the same slope, and point-slope form anchors the line.
  2. Distribute: y − 7 = 3x − 3; add 7: y = 3x + 4.Multiplication applies to both bracket terms; adding 7 isolates y.
  3. Plug in (1, 7): 3 × 1 + 4 = 7.The chosen point must fit the final equation.
  4. Compare intercepts 4 and 6.Same slope and different intercepts confirm distinct parallel lines.
Answer
y = 3x + 4.
Check Substitution in b = y − mx gives b = 7 − 3 × 1 = 4, matching the point-slope route.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: the parallel answer through (1, 7) is y = 3x + 6.
At x = 1 that equation gives 9, so it misses the required height 7.
✓ Instead: Keep slope 3, but solve for the new intercept 4.
Tips and tricks
  • Tip: Same slope does not mean same intercept. Use the point to find the intercept.
.2Perpendicular through a point

Turn the direction through a right angle, then use the address to position the turned line. For an ordinary nonzero slope, the direction change is flip and sign. The supplied point fixes the intercept after that change.

  • Rule: Use the negative reciprocal, then solve for the new intercept.
  • Rule: Perpendicular to y = c through (p, q) is x = p.
  • Rule: Perpendicular to x = c through (p, q) is y = q.
−224−2246(0, 1)(3, 0)(3, 0)
The new line passes through the marked point; compare its slope with the dashed given line.
Reminder
  • Negative reciprocal. 3 = 31 becomes −13.
  • Signs. (−13) × (−3) = 1 because two negative factors give a positive product.
The same idea, five ways
Say it

Say: turn the direction, then fit the address.

Write it

The new line makes a right angle and passes through the required point.

In math
  • mnew = −1mgiven
  • y − y1 = mnew(x − x1)
Like

Turn a street into a cross street that passes through one location.

See it
−224−2246(0, 1)(3, 0)(3, 0)
The new line passes through the marked point; compare its slope with the dashed given line.
Worked exampleThe verified perpendicular line

You need a line making a right angle with f(x) = 3x + 3 and passing through (3, 0).

−224−2246(0, 1)(3, 0)(3, 0)
The new line passes through the marked point; compare its slope with the dashed given line.
  1. The given slope is 3 = 31, so the new slope is −13.Flip the fraction and reverse its sign.
  2. Write y − 0 = −13(x − 3).Point-slope form includes the supplied point.
  3. Distribute: y = −13x + 1.The product (−13) × (−3) is 1.
  4. Plug in x = 3: y = −1 + 1 = 0; check 3 × (−13) = −1.This verifies both the point and the right-angle slope relationship.
Answer
y = −13x + 1.
Check The alternative intercept calculation gives b = 0 − (−13 × 3) = 1.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: use slope −3 to be perpendicular to slope 3.
The product is −9, not −1.
✓ Instead: Use slope −13, whose product with 3 is −1.
Tips and tricks
  • Tip: Check the slope product and the point separately. Either condition can fail on its own.
Strategy: step by step
  1. 1. Find the original slope from its equation or from two points.
  2. 2. Choose the new slope: same for parallel, negative reciprocal for perpendicular.
  3. 3. Insert the new slope and the required point in point-slope form, or use b = y1 − mx1.
  4. 4. Expand and isolate y if slope-intercept form is wanted.
  5. 5. Plug the supplied point into the final equation and check the slope relationship.
Strategy
Strategy: write the requested related line
1
Is the given line supplied by two points?
YesCalculate its rise and run first. A zero run identifies a vertical line; zero rise with nonzero run identifies a horizontal line.
NoRead its slope after isolating y, or recognize x = c as vertical.
↓
2
Is the given line horizontal or vertical?
YesKeep its direction for parallel or swap directions for perpendicular, then set the required point's fixed coordinate. This completes the construction; do not take a reciprocal.
NoContinue using the original finite nonzero slope.
↓
3
Is the requested nonvertical line parallel?
YesKeep the original slope, then use the point to find b. Check for a different intercept if the line must be distinct.
NoTake the negative reciprocal of the original slope, then use the point to find b.
  1. Find the given slope.
  2. Decide same direction or right-angle direction.
  3. Choose the new slope or the appropriate horizontal/vertical equation.
  4. Anchor the new line at the supplied point.
  5. Check the point and relationship in the original conditions.
Worked exampleA perpendicular line from two given points

You need a line perpendicular to the line through (−2, 6) and (4, 5), and the new line must pass through (4, 5).

246−4−224681012(0, −19)(3.17, 0)required meeting point
The new line passes through the marked point; compare its slope with the dashed given line.
  1. Find the original slope: 5−64−(−2) = −16 = −16.Subtract the same point order for outputs and inputs; the run is 6.
  2. The negative reciprocal of −16 is 6.Flipping makes −6, then reversing the sign makes 6.
  3. Write y − 5 = 6(x − 4).The chosen slope and required point determine the new line.
  4. Distribute: y − 5 = 6x − 24. Add 5: y = 6x − 19.Distribution and balanced addition put the answer in slope-intercept form.
  5. Plug in (4, 5): 6 × 4 − 19 = 24 − 19 = 5. Also (−16) × 6 = −1.The equation must satisfy the point and the perpendicular condition.
Answer
y = 6x − 19.
Check The original line's intercept is 5 − (−16 × 4) = 173, so it is y = −16x + 173. At x = −2 it gives 13 + 173 = 6; at 4 it gives 5. Its slope product with the new line is −1.
Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.
Rung 1Rung 1: parallel with a whole-number slope

You need the parallel line to y = 3x + 6 through (1, 7).

224681012(0, 4)(−1.33, 0)point
The new line passes through the marked point; compare its slope with the dashed given line.
  1. Keep m = 3 and substitute 7 = 3 × 1 + b.Same slope supplies the direction; the point supplies placement.
  2. Subtract 3 to find b = 4. Plug back in: 3 + 4 = 7.This finds and checks the intercept.
  3. Write y = 3x + 4; intercepts 4 and 6 differ.Distinct parallel lines have equal slopes and different intercepts.
Answer
y = 3x + 4.
Check Point-slope form y − 7 = 3(x − 1) expands to the same equation.
Rung 2Rung 2: parallel with a fractional slope

You need a line parallel to y = 23x − 1 through (3, 4).

−4−2246−6−4−2246810(0, 2)(−3, 0)point
The new line passes through the marked point; compare its slope with the dashed given line.
  1. Keep m = 23 and write 4 = 23 × 3 + b = 2 + b.The point must fit the new slope.
  2. Subtract 2: b = 2. Plug back in: 23 × 3 + 2 = 4.This determines and verifies the starting output.
  3. Write y = 23x + 2.Same slope and different intercepts give the parallel relationship.
Answer
y = 23x + 2.
Check Point-slope form y − 4 = 23(x − 3) expands to y = 23x + 2.
Rung 3Rung 3: perpendicular to a positive slope

You need the perpendicular line to y = 3x + 3 through (3, 0).

−4−2246−6−4−2246810(0, 1)(3, 0)point
The new line passes through the marked point; compare its slope with the dashed given line.
  1. Choose m = −13, then use 0 = −13 × 3 + b = −1 + b.The negative reciprocal turns the direction through a right angle.
  2. Add 1 to find b = 1. Plug back in: −1 + 1 = 0.This finds the placement that fits the supplied point.
  3. Write y = −13x + 1.The slope and intercept have both been determined.
Answer
y = −13x + 1.
Check Point-slope form gives the same equation; the slope product 3 × (−13) is −1.
Rung 4Rung 4: perpendicular to a negative slope

You need the perpendicular line to y = −25x + 4 through (2, −1).

−4−2246−6−4−2246810(0, −6)(2.4, 0)point
The new line passes through the marked point; compare its slope with the dashed given line.
  1. The negative reciprocal is 52. Substitute −1 = 52 × 2 + b = 5 + b.A negative original slope gives a positive perpendicular slope.
  2. Subtract 5 to obtain b = −6. Plug back in: 5 − 6 = −1.The required point selects and verifies the intercept.
  3. Write y = 52x − 6.The chosen direction is now positioned correctly.
Answer
y = 52x − 6.
Check The slope product (−25) × 52 = −1, and point-slope form y + 1 = 52(x − 2) expands to the same result.
Rung 5Rung 5: perpendicular to a two-point line

You need the perpendicular line to the line through (−2, 6) and (4, 5), passing through (4, 5).

246−4−224681012(0, −19)(3.17, 0)point
The new line passes through the marked point; compare its slope with the dashed given line.
  1. Original rise = 5 − 6 = −1 and run = 4 − (−2) = 6. Thus mold = −16.Both changes measure the trip from the first supplied point to the second.
  2. Flip the old slope to −6, then reverse its sign: mnew = 6.The negative reciprocal makes the slope product −1.
  3. Use y − 5 = 6(x − 4).This anchors the requested slope at the supplied point (4, 5).
  4. Distribute: y − 5 = 6x − 24.The multiplier 6 reaches both x and −4.
  5. Add 5 to both sides: y = 6x − 24 + 5 = 6x − 19.This leaves y alone and finds the new line's intercept.
  6. Substitute the point: 6 × 4 − 19 = 5. Also (−16) × 6 = −1.Both the placement condition and the right-angle condition must be true.
Answer
y = 6x − 19.
Check Direct intercept substitution gives b = 5 − 24 = −19, matching point-slope form.
Rung 6Rung 6: horizontal and vertical exceptions

You need the perpendicular line to y = 1 through (−4, 9), and the perpendicular line to x = 7 through that same point.

Given point: (−4, 9)
Against y = 1: x = −4
Against x = 7: y = 9
The direction swaps and the point supplies the fixed coordinate.
  1. Perpendicular to the horizontal y = 1 is vertical. Through (−4, 9), keep x fixed at −4.Horizontal and vertical directions make a square corner; no reciprocal of zero is needed.
  2. Perpendicular to the vertical x = 7 is horizontal. Through (−4, 9), keep y fixed at 9.The undefined vertical slope is handled by direction instead of division.
  3. Check the required point has x = −4 and y = 9.Both constructed equations must contain that point.
Answer
  • Perpendicular to y = 1: x = −4.
  • Perpendicular to x = 7: y = 9.
Check The meeting points are (−4, 1) and (7, 9); each pair is horizontal against vertical, so each angle is 90°.
Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: changing the slope to 6 while keeping the old intercept automatically makes the line pass through (4, 5).
Slope controls direction, while the intercept controls placement. One condition does not ensure the other.
✓ Instead: Use b = 5 − 6 × 4 = −19 and verify 6 × 4 − 19 = 5.
Tips and tricks
  • Tip: Underline the supplied point; it belongs to the requested line.
  • Tip: Use point-slope form to avoid losing the point while expanding.
  • Tip: If the point is on the old line and distinct parallel lines are requested, no such line exists; the same-slope construction is coincident.
Trap. Using the old slope in a perpendicular problem, or using the old intercept after changing the slope. Choose the direction first, then use the new point to find the new intercept.