Gradients of Parallel and Perpendicular Lines

Equally steep, or multiplying to minus one.

Parallel lines have equal gradients

The lines y = 2x and y = 2x + 3 both rise 2 for every 1 across. At every value of x, the second line is (2x + 3) − 2x = 3 above the first. The gap between them is 3 everywhere, so they never meet: they are parallel.

The algebra says the same. Where two lines meet, their y-values are equal, so y = 2x and y = 2x + 3 would meet where 2x = 2x + 3. Subtracting 2x from both sides leaves 0 = 3, which is never true, so no point is on both lines.

Lines with different gradients always meet. For y = 2x and y = 3x + 1, the gap is (3x + 1) − 2x = x + 1. It changes by 1 for each step of 1 across, so somewhere it is 0: at x = −1, where both lines pass through (−1, −2). In general the gap changes by the difference of the two gradients for each step across, so it reaches 0 somewhere unless that difference is 0. So two lines that are not vertical are parallel exactly when their gradients are equal. If their y-intercepts are equal as well, they are the same line.

xy

The gold line is y = 2x and the white line is y = 2x + 3. Both rise 2 for every 1 across, and the white line is 3 above the gold one at every x.

A quarter turn of the slope triangle

The line y = 2x goes 1 across and 2 up. Turn that step a quarter turn clockwise: what pointed across now points down, and what pointed up now points across. The step becomes 2 across and 1 down.

The turned step has turned through 90°, so the line it makes is perpendicular to y = 2x. Its gradient is −1 ÷ 2 = −1/2.

The same works for any gradient m. A line with gradient m goes 1 across and m up. The quarter turn makes that m across and 1 down, a gradient of −1/m. So the perpendicular gradient is the original turned upside down, with its sign changed.

xy1221

The gold line y = 2x steps 1 across and 2 up, from (3, 6) to (4, 8). The white line y = −x/2 + 5 steps 2 across and 1 down, from (4, 3) to (6, 2): the same step, turned a quarter turn. The two lines cross at (2, 4), at a right angle.

The product is −1

Multiply a gradient by its perpendicular gradient: m × (−1/m) = −1. For y = 2x and y = −x/2 + 5, 2 × (−1/2) = −1. So two lines are perpendicular when their gradients multiply to −1: m₁ × m₂ = −1.

To find a perpendicular gradient, turn the gradient upside down and change its sign. 3/4 becomes −4/3; −5, which is −5/1, becomes 1/5; and 1 becomes −1. Check each by multiplying: 3/4 × (−4/3) = −1, −5 × 1/5 = −1 and 1 × (−1) = −1.

When neither line is level, one of two perpendicular lines rises and the other falls, which is why their gradients have opposite signs.

Any pair works

A line with gradient 3 and a line with gradient −1/3 cross at a right angle, since 3 × (−1/3) = −1. The steep line goes 1 across and 3 up; the other goes 3 across and 1 down.

xy

The gold line y = 3x and the white line y = −x/3 + 3 cross at a right angle: 3 × (−1/3) = −1.

Level and vertical lines

A level line such as y = 2 and a vertical line such as x = 5 are perpendicular. The product rule cannot show it: the level line has gradient 0, and the vertical line has no gradient to multiply. This pair is the one exception.

Testing lines written as Ax + By = C

To compare lines, get y on its own in each equation first. For 2x + 3y = 6, subtract 2x and divide by 3: y = −2x/3 + 2, so the gradient is −2/3. For 3x − 2y = 4, subtract 3x and divide by −2: y = 3x/2 − 2, so the gradient is 3/2. The product is −2/3 × 3/2 = −1, so the lines are perpendicular.

For y = 4x − 1 and 8x − 2y = 5, the second rearranges to y = 4x − 5/2. Both gradients are 4 and the intercepts differ, so the lines are parallel.

A parallel or perpendicular line through a point

The line through (1, 6) parallel to y = 3x − 2 has the same gradient, 3. By point-slope form it is y − 6 = 3(x − 1), so y − 6 = 3x − 3 and y = 3x + 3. Check: 3 × 1 + 3 = 6.

The line through (4, 1) perpendicular to y = 2x + 1 has gradient −1/2. It is y − 1 = −1/2 (x − 4), so y − 1 = −x/2 + 2 and y = −x/2 + 3. Check: −4/2 + 3 = 1.

The usual mistakes

Turning the gradient over but keeping its sign. A line perpendicular to gradient 2 does not have gradient 1/2: 2 × 1/2 = 1, not −1, and two rising lines cannot cross at a right angle. The gradient is −1/2.

Changing the sign without turning it over. −2 × 2 = −4, not −1. Both moves are needed.

Using the perpendicular gradient for a parallel line. Parallel lines have the same gradient: parallel to gradient 2 is gradient 2.

Reading the gradient straight from Ax + By = C. In 2x + 3y = 6 the gradient is not 2; with y on its own, the equation is y = −2x/3 + 2, and the gradient is −2/3.

Worked example: The Shortest Track from a Farmhouse to a Straight Road

Question On a map with a grid in kilometers, a straight road passes through (3, 6) and (9, 14). A farmhouse is at H(10, 7). The farmer wants the shortest possible straight track from the farmhouse to the road. (a) Find the equation of the line along which the track must run. (b) Find the point F where the track meets the road, and the length of the track.

  1. 1.Find the equation of the road. Its gradient is m = 14 − 69 − 3 = 86 = 43. Use the point (3, 6): y − 6 = 43(x − 3), so y − 6 = 43x − 4 and y = 43x + 2.

    4812160481216km east, xkm north, y(3, 6)(9, 14)H(10, 7)road: m = (14 − 6)/(9 − 3) = 8/6 = 4/3y − 6 = 4/3 (x − 3), so y = 4/3 x + 2
    4812160481216km east, xkm north, y(3, 6)(9, 14)H(10, 7)road: m = (14 − 6)/(9 − 3) = 8/6 = 4/3y − 6 = 4/3 (x − 3), so y = 4/3 x + 2
    The road has gradient 14 − 69 − 3 = 43. Through (3, 6) its equation is y − 6 = 43(x − 3), which simplifies to y = 43x + 2.

  2. 2.The shortest track is perpendicular to the road. Perpendicular gradients multiply to −1, so the gradient of the track is the negative reciprocal, −34. Check: 43 × (−34) = −1.

    4812160481216km east, xkm north, y(3, 6)(9, 14)H(10, 7)the shortest track is perpendicular to the road4/3 × (−3/4) = −1, so its gradient is −3/4
    4812160481216km east, xkm north, y(3, 6)(9, 14)H(10, 7)the shortest track is perpendicular to the road4/3 × (−3/4) = −1, so its gradient is −3/4
    The shortest track is perpendicular to the road. Perpendicular gradients multiply to −1, so the gradient of the track is −34.

  3. 3.(a) The track passes through H(10, 7): y − 7 = −34(x − 10), so y − 7 = −34x + 7.5 and y = −34x + 14.5.

    4812160481216km east, xkm north, y(3, 6)(9, 14)H(10, 7)y − 7 = −3/4 (x − 10)y = −3/4 x + 7.5 + 7, so y = −3/4 x + 14.5
    4812160481216km east, xkm north, y(3, 6)(9, 14)H(10, 7)y − 7 = −3/4 (x − 10)y = −3/4 x + 7.5 + 7, so y = −3/4 x + 14.5
    (a) Through H(10, 7) the track runs along y − 7 = −34(x − 10), which simplifies to y = −34x + 14.5.

  4. 4.At F both equations hold: 43x + 2 = −34x + 14.5. Multiply both sides by 12: 16x + 24 = −9x + 174, so 25x = 150 and x = 6. Then y = 43 × 6 + 2 = 10.

    4812160481216km east, xkm north, yF(6, 10)(3, 6)(9, 14)H(10, 7)4/3 x + 2 = −3/4 x + 14.5; multiply both sides by 1216x + 24 = −9x + 174, so x = 6 and y = 10
    4812160481216km east, xkm north, yF(6, 10)(3, 6)(9, 14)H(10, 7)4/3 x + 2 = −3/4 x + 14.5; multiply both sides by 1216x + 24 = −9x + 174, so x = 6 and y = 10
    At F both equations hold: 43x + 2 = −34x + 14.5. Multiply both sides by 12: 16x + 24 = −9x + 174, so x = 6 and y = 10.

  5. 5.From F(6, 10) to H(10, 7) the difference in x is 4 and the difference in y is −3, so FH = √42 + (−3)2 = √25 = 5.

    4812160481216km east, xkm north, y5 kmF(6, 10)(3, 6)(9, 14)H(10, 7)from F to H: 4 across and 3 downFH2= 42+ 32= 25, so FH =√25= 5 km
    4812160481216km east, xkm north, y5 kmF(6, 10)(3, 6)(9, 14)H(10, 7)from F to H: 4 across and 3 downFH2= 42+ 32= 25, so FH =√25= 5 km
    From F(6, 10) to H(10, 7) the map goes 4 across and 3 down, so FH = √42 + 32 = 5 km.

  6. 6.(b) The track meets the road at F(6, 10) and is 5 km long. Check that F is on the track: −34 × 6 + 14.5 = −4.5 + 14.5 = 10.

    4812160481216km east, xkm north, y5 kmF(6, 10)(3, 6)(9, 14)H(10, 7)the track meets the road at F(6, 10) and is 5 km longcheck: −3/4 × 6 + 14.5 = 10
    4812160481216km east, xkm north, y5 kmF(6, 10)(3, 6)(9, 14)H(10, 7)the track meets the road at F(6, 10) and is 5 km longcheck: −3/4 × 6 + 14.5 = 10
    (b) The track meets the road at F(6, 10) and is 5 km long. The point F satisfies the equation of the track as well.

Answer: (a) y = −34x + 14.5; (b) F(6, 10), and the track is 5 km long

Common mistakes

  • Building the track due west from the farmhouse, along y = 7. It meets the road where 43x + 2 = 7, at x = 3.75, so it is 10 − 3.75 = 6.25 km long. The perpendicular track is 5 km long, which is shorter.
  • Using 34 for the gradient of the track, without the negative sign. Then 43 × 34 = 1, not −1. One of two perpendicular lines slopes upward and the other slopes downward, so their gradients have opposite signs.

More quadratic graphs and coordinate geometry problems, worked step by step →

Worked example: Four Fence Posts Shown to Be the Corners of a Rectangle

Question Four fence posts are marked on a plan with a grid in meters: P(5, 2), Q(13, 8), R(10, 12) and S(2, 6). (a) Show that PQRS is a rectangle. (b) Find the length of fencing needed to go all the way round PQRS, and the area that the fence encloses.

  1. 1.Find the gradients of one pair of opposite sides. For PQ: 8 − 213 − 5 = 68 = 34. For SR: 12 − 610 − 2 = 68 = 34. The gradients are equal, so PQ is parallel to SR.

    48120481216meters east, xmeters north, y3/43/4P(5, 2)Q(13, 8)R(10, 12)S(2, 6)gradient of PQ = (8 − 2)/(13 − 5) = 6/8 = 3/4gradient of SR = (12 − 6)/(10 − 2) = 6/8 = 3/4
    48120481216meters east, xmeters north, y3/43/4P(5, 2)Q(13, 8)R(10, 12)S(2, 6)gradient of PQ = (8 − 2)/(13 − 5) = 6/8 = 3/4gradient of SR = (12 − 6)/(10 − 2) = 6/8 = 3/4
    The gradient of PQ is 8 − 213 − 5 = 34 and the gradient of SR is 12 − 610 − 2 = 34. Equal gradients mean that PQ is parallel to SR.

  2. 2.Find the gradients of the other pair. For QR: 12 − 810 − 13 = 4−3 = −43. For PS: 6 − 22 − 5 = 4−3 = −43. So QR is parallel to PS, and PQRS is a parallelogram.

    48120481216meters east, xmeters north, y3/43/4−4/3−4/3P(5, 2)Q(13, 8)R(10, 12)S(2, 6)gradient of QR = (12 − 8)/(10 − 13) = −4/3gradient of PS = (6 − 2)/(2 − 5) = −4/3
    48120481216meters east, xmeters north, y3/43/4−4/3−4/3P(5, 2)Q(13, 8)R(10, 12)S(2, 6)gradient of QR = (12 − 8)/(10 − 13) = −4/3gradient of PS = (6 − 2)/(2 − 5) = −4/3
    The gradients of QR and PS are both 4−3 = −43, so these sides are parallel as well and PQRS is a parallelogram.

  3. 3.(a) The gradients of PQ and QR multiply to 34 × (−43) = −1, so the angle at Q is a right angle. A parallelogram with a right angle is a rectangle.

    48120481216meters east, xmeters north, y3/4−4/3P(5, 2)Q(13, 8)R(10, 12)S(2, 6)3/4 × (−4/3) = −1, so the angle at Q is 90 dega parallelogram with a right angle is a rectangle
    48120481216meters east, xmeters north, y3/4−4/3P(5, 2)Q(13, 8)R(10, 12)S(2, 6)3/4 × (−4/3) = −1, so the angle at Q is 90 dega parallelogram with a right angle is a rectangle
    (a) 34 × (−43) = −1, so PQ is perpendicular to QR. A parallelogram with a right angle is a rectangle.

  4. 4.Find the lengths of two neighboring sides: PQ = √82 + 62 = √100 = 10 m and QR = √(−3)2 + 42 = √25 = 5 m.

    48120481216meters east, xmeters north, y10 m5 mP(5, 2)Q(13, 8)R(10, 12)S(2, 6)PQ2= 82+ 62= 100, so PQ =√100 = 10 mQR2= 32+ 42= 25, so QR =√25= 5 m
    48120481216meters east, xmeters north, y10 m5 mP(5, 2)Q(13, 8)R(10, 12)S(2, 6)PQ2= 82+ 62= 100, so PQ =√100 = 10 mQR2= 32+ 42= 25, so QR =√25= 5 m
    The distance formula gives PQ = √82 + 62 = 10 m and QR = √32 + 42 = 5 m.

  5. 5.(b) The fencing is the perimeter, 2 × (10 + 5) = 30 m. The area enclosed is 10 × 5 = 50 m2.

    48120481216meters east, xmeters north, y10 m5 m50 m2P(5, 2)Q(13, 8)R(10, 12)S(2, 6)fencing: 2 × (10 + 5) = 30 marea: 10 × 5 = 50 m2
    48120481216meters east, xmeters north, y10 m5 m50 m2P(5, 2)Q(13, 8)R(10, 12)S(2, 6)fencing: 2 × (10 + 5) = 30 marea: 10 × 5 = 50 m2
    (b) The fencing is the perimeter, 2 × (10 + 5) = 30 m, and the area enclosed is 10 × 5 = 50 m2.

Answer: (a) the gradients of PQ and SR are both 34, the gradients of QR and PS are both −43, and 34 × (−43) = −1; (b) 30 m of fencing, and an area of 50 m2

Common mistakes

  • Stopping once the opposite sides are shown to be parallel. That proves only that PQRS is a parallelogram, which may lean over. A rectangle also needs a right angle, which is shown by two gradients that multiply to −1.
  • Finding the area as the width of the plan times its height, 11 × 10 = 110 m2. The rectangle is tilted on the grid, so its sides are not along the grid lines. Their lengths, 10 m and 5 m, come from the distance formula.

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