Gradient from Two Points

Rise over run, without the equation.

Two points and no equation

Only one straight line passes through the two points (1, 2) and (4, 8). How steep is it? There is no equation to read a gradient from, so the steepness has to come from the coordinates themselves.

The gradient of a line is how far it goes up for each 1 it goes across. Between two points, that is the rise divided by the run: the change in y divided by the change in x.

Rise over run

From (1, 2) to (4, 8), x goes from 1 to 4, so the run is 4 − 1 = 3. And y goes from 2 to 8, so the rise is 8 − 2 = 6.

A rise of 6 spread over a run of 3 is 6 ÷ 3 = 2 up for each 1 across. So the gradient is m = 2, the same m as in y = mx + c.

xyrun 3rise 6

From (1, 2) to (4, 8) the run is 3 and the rise is 6, so the gradient is 6 ÷ 3 = 2.

Any two points give the same gradient

The point (2, 4) is on the same line. From (1, 2) to (2, 4) the run is 1 and the rise is 2, and 2 ÷ 1 = 2. From (2, 4) to (4, 8) the run is 2 and the rise is 4, and 4 ÷ 2 = 2.

The two slope triangles are different sizes but have the same angles. Each has a right angle between its run and its rise, and each has its sloping side on the same straight line, which meets every horizontal at the same angle. So the triangles are similar, and in similar triangles the sides are in the same ratio: the rise divided by the run is the same in both.

This is true for any two points on the line. The gradient belongs to the line, not to the points chosen to measure it.

xy1224

Two slope triangles on one line. The small one, from (1, 2) to (2, 4), has run 1 and rise 2. The large one, from (2, 4) to (4, 8), has run 2 and rise 4. Both give a gradient of 2.

The formula

Call the two points (x₁, y₁) and (x₂, y₂). The small 1 and 2 below the letters only say which point is first and which is second. Then the rise is y₂ − y₁ and the run is x₂ − x₁, so m = (y₂ − y₁) / (x₂ − x₁).

Either point can be the first one, as long as the same order is used on the top and the bottom. Taking (4, 8) first gives (2 − 8) ÷ (1 − 4) = −6 ÷ (−3) = 2: both differences change sign, and the gradient does not change.

Mixing the orders does change it. (8 − 2) ÷ (1 − 4) = 6 ÷ (−3) = −2 has the right size and the wrong sign, because the top went from the first point to the second and the bottom went back.

A line that goes downhill

From (1, 7) to (4, 1), the rise is 1 − 7 = −6 and the run is 4 − 1 = 3, so the gradient is −6 ÷ 3 = −2. The line drops 2 for each 1 across.

With the left-hand point taken first, the run is positive. So the sign of the gradient comes from the rise: positive when y goes up, negative when y goes down. A gradient of −2 is just as steep as a gradient of 2; the minus sign says which way the line slopes.

Negative coordinates and fractional gradients

From (−2, 1) to (4, 5), the run is 4 − (−2) = 4 + 2 = 6. Subtracting a negative coordinate adds its size, which is right: from −2 to 0 is 2 across, and from 0 to 4 is 4 more. The rise is 5 − 1 = 4.

So m = 4 / 6 = 2/3. A gradient does not have to be a whole number. This one means 2 up for every 3 across, or 2/3 of a unit up for each 1 across.

xyrun 6rise 4

From (−2, 1) to (4, 5) the run is 4 − (−2) = 6 and the rise is 4, so the gradient is 4/6 = 2/3.

Level lines and vertical lines

Through (1, 3) and (5, 3) the rise is 3 − 3 = 0, so the gradient is 0 ÷ 4 = 0. Two points with the same y-coordinate lie on a level line, and a level line has gradient 0.

Through (2, 1) and (2, 6) the run is 2 − 2 = 0, and the gradient would be 5 ÷ 0. Division by 0 has no answer, so a vertical line has no gradient. Its equation is x = 2, which is not of the form y = mx + c.

Are three points on one line?

Because a line has the same gradient between any two of its points, gradients can test whether three points lie on one straight line. Points that do are called collinear.

Take A(1, 1), B(3, 4) and C(7, 10). From A to B the gradient is (4 − 1) ÷ (3 − 1) = 3/2. From B to C it is (10 − 4) ÷ (7 − 3) = 6/4 = 3/2. The two gradients are equal and the two pieces share the point B, so A, B and C are collinear.

Now take D(7, 9) instead of C. From B to D the gradient is (9 − 4) ÷ (7 − 3) = 5/4, which is not 3/2, so A, B and D are not on one line, even though D is only 1 below C.

The usual mistakes

Subtracting in different orders on the top and the bottom. For (1, 2) and (4, 8), (8 − 2) ÷ (1 − 4) gives −2, not 2. Take both differences from the same first point.

Dividing the run by the rise. 3 ÷ 6 = 1/2 is the run for each 1 up. The gradient is the rise for each 1 across: 6 ÷ 3 = 2.

Losing the sign of a negative coordinate. From (−2, 1) to (4, 5) the run is 4 − (−2) = 6, not 4 − 2 = 2.

Worked example: A Wheelchair Ramp Checked Against the Steepest Gradient the Building Code Allows

Question A surveyor draws the side view of a wheelchair ramp on axes, with x m the horizontal distance from a gate post and y m the height above the pavement. The ramp is a straight line from its foot at (3, 0.1) to the door at (9, 0.7). The building code says that the gradient of a ramp must not be more than 112. (a) Find the gradient of the ramp and decide whether the code allows it. (b) The door cannot be moved. Where must the foot of the ramp be, still at a height of 0.1 m, for the gradient to be exactly 112?

  1. 1.Find the rise and the run from the two points. The rise is 0.7 − 0.1 = 0.6 m and the run is 9 − 3 = 6 m, so the gradient is 0.66 = 0.1 = 110.

    0.510246810meters from the gate post, xheight above the pavement (m), yrun 6rise 0.6(3, 0.1)(9, 0.7)rise = 0.7 − 0.1 = 0.6 m, run = 9 − 3 = 6 mgradient = 0.6/6 = 0.1 = 1/10
    0.510246810meters from the gate post, xheight above the pavement (m), yrun 6rise 0.6(3, 0.1)(9, 0.7)rise = 0.7 − 0.1 = 0.6 m, run = 9 − 3 = 6 mgradient = 0.6/6 = 0.1 = 1/10
    The rise is 0.7 − 0.1 = 0.6 m and the run is 9 − 3 = 6 m, so the gradient is 0.66 = 110. The height axis is stretched so that the rise can be seen.

  2. 2.(a) Compare 110 with 112. The numerators are equal and 10 is less than 12, so 110 is the larger fraction. The ramp is steeper than the code allows.

    0.510246810meters from the gate post, xheight above the pavement (m), yrun 6rise 0.6(3, 0.1)(9, 0.7)1/10 is more than 1/12the ramp is steeper than the code allows
    0.510246810meters from the gate post, xheight above the pavement (m), yrun 6rise 0.6(3, 0.1)(9, 0.7)1/10 is more than 1/12the ramp is steeper than the code allows
    (a) With equal numerators, the smaller denominator gives the larger fraction, so 110 is more than 112. The ramp is steeper than the code allows.

  3. 3.The door stays at (9, 0.7) and the foot stays at a height of 0.1 m, so the rise is still 0.6 m. For a gradient of 112 the run must be 12 times the rise: 12 × 0.6 = 7.2 m.

    0.510246810meters from the gate post, xheight above the pavement (m), y(9, 0.7)run 7.2rise 0.6the rise is still 0.6 ma gradient of 1/12 needs a run of 12 × 0.6 = 7.2 m
    0.510246810meters from the gate post, xheight above the pavement (m), y(9, 0.7)run 7.2rise 0.6the rise is still 0.6 ma gradient of 1/12 needs a run of 12 × 0.6 = 7.2 m
    The door and the height of the foot do not change, so the rise is still 0.6 m. A gradient of 112 needs a run of 12 × 0.6 = 7.2 m.

  4. 4.The foot is 7.2 m before the door, so its x-coordinate is 9 − 7.2 = 1.8.

    0.510246810meters from the gate post, xheight above the pavement (m), y(9, 0.7)run 7.2rise 0.6(1.8, 0.1)the foot is 7.2 m before the doorx = 9 − 7.2 = 1.8
    0.510246810meters from the gate post, xheight above the pavement (m), y(9, 0.7)run 7.2rise 0.6(1.8, 0.1)the foot is 7.2 m before the doorx = 9 − 7.2 = 1.8
    The foot is 7.2 m before the door, so its x-coordinate is 9 − 7.2 = 1.8.

  5. 5.(b) The foot of the ramp must be at (1.8, 0.1), which is 3 − 1.8 = 1.2 m further from the door than it is now. Check: 0.67.2 = 672 = 112.

    0.510246810meters from the gate post, xheight above the pavement (m), y(9, 0.7)run 7.2rise 0.6(1.8, 0.1)the foot must be at (1.8, 0.1)check: 0.6/7.2 = 6/72 = 1/12
    0.510246810meters from the gate post, xheight above the pavement (m), y(9, 0.7)run 7.2rise 0.6(1.8, 0.1)the foot must be at (1.8, 0.1)check: 0.6/7.2 = 6/72 = 1/12
    (b) The foot of the ramp must be at (1.8, 0.1), which is 1.2 m further from the door. Then the gradient is 0.67.2 = 112.

Answer: (a) the gradient is 110, which is more than 112, so the code does not allow it; (b) at (1.8, 0.1), which makes the run 7.2 m

Common mistakes

  • Dividing the run by the rise, which gives 10 and makes the ramp seem far too steep. The gradient is the change in y divided by the change in x: the height gained for each meter along the ground.
  • Deciding that 110 is less than 112 because 10 is less than 12. When the numerators are equal, the fraction with the smaller denominator is the larger one: a tenth of a meter is more than a twelfth of a meter.

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