8C - Gradients of straight lines

8C - Gradients of straight lines

A line can rise gently, fall steeply or stand vertically. You will describe its direction using coordinate changes, distinguish zero gradient from undefined gradient and decide whether three points lie on one straight line.

Change in y over change in x

If a line rises 33 units while moving 66 units to the right, what number describes its steepness?

The rise per unit of horizontal movement is 3/6=123/6=\tfrac12. Doubling both changes to 66 and 1212 gives the same quotient. Right triangles drawn along one straight line are similar, so this ratio is constant wherever you measure it.

[DIAGRAM: asset_name: Positive and negative gradient; asset_slug: gradient-changes; caption: Equal horizontal changes give opposite gradients when the vertical change is reversed; description: Two equal-scale coordinate panels show a rising segment from minus 2 comma 1 to 4 comma 4 with run 6 and rise 3, and a falling segment from minus 2 comma 4 to 4 comma 1 with run 6 and rise minus 3. Their gradients are one-half and minus one-half.]
Equal horizontal changes give opposite gradients when the vertical change is reversed

Gradient

The gradient of a nonvertical straight line is the change in yy divided by the corresponding change in xx. It measures signed rise per unit horizontal change.

Read direction from left to right: a rising line has positive gradient, and a falling line has negative gradient. For example, a gradient of −2-2 means a change of −2-2 in yy for each change of 11 in xx.

Gradient from two points

For distinct points (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2), with x2≠x1x_2\ne x_1,

m=y2−y1x2−x1.m=\frac{y_2-y_1}{x_2-x_1}.

Subtract the coordinates in the same endpoint order in the numerator and denominator.

Not on the formulae sheet: learn it.

The quotient is signed: unlike the distance formula, we do not square or take absolute values of the changes.

Worked example 1: a rising line

Question. Find the gradient of the line through A(−2,1)A(-2,1) and B(4,4)B(4,4).

Plan. Divide the change in yy from AA to BB by the matching change in xx.

m=4−14−(−2)use the order B−A=36=12simplify exactly\begin{aligned} m&=\frac{4-1}{4-(-2)} &&\text{use the order }B-A\\ &=\frac36=\frac12 &&\text{simplify exactly} \end{aligned}

Answer. m=12m=\tfrac12.

Check. A horizontal change of 66 gives predicted rise 12×6=3\tfrac12\times6=3, matching the coordinates.

Keep the horizontal positions, but reverse the vertical change. This should reverse the sign of the gradient.

Worked example 2: a falling line

Question. Find the gradient through C(−2,4)C(-2,4) and D(4,1)D(4,1).

Plan. Keep the negative rise when dividing by the positive run.

m=1−44−(−2)=−36=−12\begin{aligned} m&=\frac{1-4}{4-(-2)}\\ &=\frac{-3}{6}=-\frac12 \end{aligned}

Answer. m=−12m=-\tfrac12.

Check. Reversing the endpoint order gives 4−1−2−4=3−6=−12\dfrac{4-1}{-2-4}=\dfrac3{-6}=-\tfrac12, the same gradient.

Reversing both subtractions changes both signs and preserves the quotient. Reversing just one subtraction changes the answer incorrectly.

Watch out

For the first pair, 4−1−2−4=−12\dfrac{4-1}{-2-4}=-\tfrac12 mixes endpoint orders. Use 4−14−(−2)\dfrac{4-1}{4-(-2)} or 1−4−2−4\dfrac{1-4}{-2-4}; both give 12\tfrac12.

The next endpoints are given from right to left; the same rule works.

Horizontal and vertical lines

A zero numerator and a zero denominator mean different things in a gradient calculation.

Along a horizontal line, yy does not change, so the rise is zero. Along a vertical line, xx does not change, so there is no nonzero horizontal movement by which to divide.

[DIAGRAM: asset_name: Zero and undefined gradients; asset_slug: special-gradients; caption: A horizontal line has gradient zero, while a vertical line has no finite gradient; description: Two equal-scale panels show a horizontal line y equals 3 through minus 4 comma 3 and 5 comma 3, and a vertical line x equals 2 through 2 comma minus 5 and 2 comma 4. The corresponding changes are rise 0 run 9 and rise 9 run 0.]
A horizontal line has gradient zero, while a vertical line has no finite gradient

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