E4.2 - Gradient from two points (Extended)

E4.2 - Gradient from two points (Extended)

A line passes through (−2,−1)(-2,-1) and (4,2)(4,2), but no grid is supplied. Can you still measure its gradient? E4.2 turns the rise-and-run idea into a coordinate calculation, explains consistent subtraction and uses a known gradient to recover a missing coordinate.

Coordinate differences

On a grid, the horizontal change from x=1x=1 to x=7x=7 is 7−1=67-1=6. The vertical change from y=2y=2 to y=5y=5 is 5−2=35-2=3. Subtraction calculates the same two changes without requiring a drawn triangle.

Gradient from two points

For distinct points (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2) with x1≠x2x_1\ne x_2,

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

The small subscripts label the first and second points; they do not mean multiplication. Not on the supplied formula list for Papers 1–4: learn it.

Keep each point's coordinates together when assigning labels. The numerator compares vertical coordinates, and the denominator compares horizontal coordinates. Use brackets when subtracting a negative coordinate.

Worked example 1: positive coordinates

Question. Calculate the gradient of the line through (1,2)(1,2) and (7,5)(7,5).

Plan. Subtract the first point's coordinates from the second point's, then divide vertical change by horizontal change.

y2−y1=5−2=3x2−x1=7−1=6m=36=12\begin{aligned} y_2-y_1&=5-2=3\\ x_2-x_1&=7-1=6\\ m&=\frac36=\frac12 \end{aligned}

Answer. 12\frac12.

Check. A gradient of 1/21/2 over a horizontal change of 6 predicts a rise of (1/2)×6=3(1/2)\times6=3. Starting at y=2y=2 gives y=5y=5, the other endpoint.

The second pair has the same horizontal and vertical separations, but the first point now has negative coordinates. The brackets protect the signs.

Worked example 2: subtracting negative coordinates

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

Plan. Form second-minus-first differences, including each negative sign.

y2−y1=2−(−1)=3x2−x1=4−(−2)=6m=36=12\begin{aligned} y_2-y_1&=2-(-1)=3\\ x_2-x_1&=4-(-2)=6\\ m&=\frac36=\frac12 \end{aligned}

Answer. 12\frac12, answering the opening question without a grid.

Check. Moving 6 units right at a rise of half a unit per unit gives a rise of 3: −1+3=2-1+3=2. The point's negative starting height does not make the gradient negative.

If the two yy-coordinates are equal, the numerator is zero and the line is horizontal. If the two xx-coordinates are equal, the denominator is zero and the distinct points determine a vertical line with undefined gradient. Two identical points do not determine a unique line at all.

Watch out

A negative coordinate does not by itself give a negative gradient. The sign comes from the changes: here both changes are positive, even though the first point lies below and left of the origin.

Use a fresh pair that crosses both axes.

Order of subtraction

You may travel from either point to the other. Reversing direction reverses both changes, so it leaves their quotient unchanged. Reversing only one difference changes the sign and gives the wrong gradient.

For non-zero x2−x1x_2-x_1, the reason is

y1−y2x1−x2=−(y2−y1)−(x2−x1)=y2−y1x2−x1.\frac{y_1-y_2}{x_1-x_2} =\frac{-(y_2-y_1)}{-(x_2-x_1)} =\frac{y_2-y_1}{x_2-x_1}.

Both minus signs cancel together. Draw a small two-column coordinate list if you find it difficult to keep the order aligned.

Worked example 3: left point to right point

Question. Find the gradient through C(−3,4)C(-3,4) and D(5,−2)D(5,-2), using D−CD-C.

Plan. Subtract both coordinates of CC from the corresponding coordinates of DD.

m=−2−45−(−3)=−68=−34\begin{aligned} m&=\frac{-2-4}{5-(-3)}\\ &=\frac{-6}{8}\\ &=-\frac34 \end{aligned}

Answer. −34-\frac34.

Check. From left to right, the height falls from 4 to −2-2, so a negative gradient is expected. A run of 8 at this gradient gives the required change −6-6.

Keep the line and reverse only the direction of travel. This tests whether subtraction order has a mathematical effect or merely changes the working.

Continue this lesson.

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