Use similar triangles to explain constant slope and derive y = mx and y = mx + b

Learning goal: Use similar triangles to explain why the slope m of a non-vertical line is the same between any two distinct points on it, then derive the equation y = mx for a line through the origin and y = mx + b for a line intercepting the vertical axis at b.

Slope triangles. Choose any two distinct points on a line in the coordinate plane that is neither vertical nor horizontal. Draw a horizontal segment from the left point and a vertical segment down or up from the right point until they meet. The result is a right triangle whose horizontal leg is the run and whose vertical leg is the rise. The slope between the two points is rise ÷ run, with the rise taken as negative when the line falls from left to right; the triangle's legs are the sizes of the run and the rise, and the sign is the same for every pair of points on the line because a line either rises everywhere or falls everywhere. A horizontal line needs no triangle: every pair of its points has rise 0, so its slope is 0 between any two points. The line must be non-vertical because on a vertical line the run between any two points is 0, the triangle collapses to a segment, and division by 0 is undefined; a vertical line has no slope.

Why every pair of points gives the same slope. Draw two slope triangles on the same line, one small and one large. Each has a right angle. Their horizontal legs are parallel (both horizontal), and the line itself cuts across both of them, so the angle each triangle makes with the line at its left vertex (the lower-left vertex for a rising line, the upper-left for a falling one) is the same in both triangles: they are corresponding angles formed by parallel lines and a transversal. Two triangles that share two angle measures are similar. In similar triangles, ratios of corresponding sides are equal, so (rise of the small triangle) ÷ (run of the small triangle) = (rise of the large triangle) ÷ (run of the large triangle). That common ratio is m. The argument uses nothing about which two points were chosen, so the slope is the same between any two distinct points on the line. Note the direction of the logic: similarity guarantees that the ratios match; it does not say the legs match. The large triangle's legs are the small triangle's legs multiplied by one scale factor, never the small legs with the same amount added to each.

A fully worked model. The points (1, 3), (4, 9) and (7, 15) lie on one line. From (1, 3) to (4, 9): run 4 − 1 = 3, rise 9 − 3 = 6, slope 6 ÷ 3 = 2. From (1, 3) to (7, 15): run 7 − 1 = 6, rise 15 − 3 = 12, slope 12 ÷ 6 = 2. The large triangle has legs 6 and 12, exactly twice the small triangle's 3 and 6: the scale factor is 2 and the triangles are similar, so the ratios agree. From (4, 9) to (7, 15): run 3, rise 6, slope 2 again. Now find where the line meets the vertical axis. Moving left from (1, 3) by a run of 1 must drop the rise by 2 (because 2 ÷ 1 = m), which lands on (0, 1). So b = 1.

Deriving y = mx for a line through the origin. Let a line pass through the origin (0, 0) with slope m. Take any other point (x, y) on it, so x is not 0. The slope triangle from the origin to (x, y) has run x − 0 = x and rise y − 0 = y. Since the slope is the same between any two points, y ÷ x = m. Multiplying both sides by x gives y = mx. The origin also satisfies the equation, since m × 0 = 0. So every point of the line satisfies y = mx, and this is the line's equation. For a proportional relationship, this is the same y = kx from the previous lesson; the constant of proportionality is the slope.

Deriving y = mx + b for a line intercepting the vertical axis at b. Now let the line cross the vertical axis at the point (0, b) and have slope m. Take any other point (x, y) on the line, with x not 0. The slope triangle from (0, b) to (x, y) has run x − 0 = x and rise y − b. Constant slope gives (y − b) ÷ x = m. Multiply both sides by x: y − b = mx. Add b to both sides: y = mx + b. The point (0, b) itself satisfies the equation, because m × 0 + b = b. In the model above, m = 2 and b = 1, so the line is y = 2x + 1. Check every point that was used: 2(1) + 1 = 3, 2(4) + 1 = 9, 2(7) + 1 = 15. All three agree, which is the test that the derivation demands.

Reading the two equations. In y = mx + b, m tells how much y changes for each increase of 1 in x, and b is the y-value where x = 0, the vertical intercept. When b = 0 the equation collapses to y = mx and the line passes through the origin. A negative m means the line falls from left to right; a slope triangle then has a negative rise, which is a drop. Slope m = 0 means every rise is 0, and the equation becomes y = b, a horizontal line.

The misconception to avoid. A common error is treating the slope as a difference rather than a ratio: reading "the line goes up 6 when it goes across 3" as slope 6, or as slope 3, instead of 6 ÷ 3 = 2. Another is assuming that a larger slope triangle means a steeper line. It does not; the larger triangle on the same line is a scaled copy of the smaller one, and the steepness, which is the ratio, is unchanged. Similar triangles are precisely the reason the size of the triangle does not matter.

Worked three-point check: The points (2, 5), (6, 11) and (10, 17) are claimed to lie on one line. Slope from the first to the second: rise 11 − 5 = 6, run 6 − 2 = 4, so m = 6 ÷ 4 = 3/2. From the second to the third: rise 17 − 11 = 6, run 10 − 6 = 4, so m = 3/2 again. From the first to the third: rise 12, run 8, m = 12 ÷ 8 = 3/2. Equal slopes confirm one line. Find b using (2, 5): 5 = (3/2)(2) + b, so 5 = 3 + b and b = 2. Equation: y = (3/2)x + 2. Check (6, 11): (3/2)(6) + 2 = 9 + 2 = 11. Check (10, 17): 15 + 2 = 17.

Worked line through the origin: A line passes through the origin and (4, −6). The slope triangle from the origin has run 4 and rise −6, so m = −6 ÷ 4 = −3/2 and the equation is y = −(3/2)x. Check the point (−2, 3): −(3/2)(−2) = 3, so it is on the line. Check (6, −9): −(3/2)(6) = −9, on the line as well.

Worked similar-triangle scaling: On a line, one slope triangle has run 2 and rise 5. A second slope triangle on the same line has run 8. Since the triangles are similar with scale factor 8 ÷ 2 = 4, its rise is 5 × 4 = 20, and both triangles give slope 5 ÷ 2 = 20 ÷ 8 = 5/2. A student who adds 6 to each leg to get run 8 and rise 11 has a triangle with slope 11 ÷ 8, which is not on this line: similarity multiplies corresponding sides, it does not add to them.

Worked intercept from slope and one point: A line has slope −2 and passes through (3, 1). Using (y − b) ÷ x = m with (3, 1): (1 − b) ÷ 3 = −2, so 1 − b = −6 and b = 7. Equation: y = −2x + 7. Check: −2(3) + 7 = 1. The line intercepts the vertical axis at (0, 7) and falls 2 units for every 1 unit to the right.

Representation task: On graph paper, draw the line through (1, 3) and (7, 15). Draw three different slope triangles on it, including one that starts at the vertical intercept (0, 1). Record run, rise and rise ÷ run for each in a table. Mark the pairs of equal angles that justify similarity and write the scale factor between each pair of triangles. Then write the equation of the line and verify it at all four labeled points.

Independent performance: Choose a nonzero slope m (a fraction or a negative number is encouraged; a slope of 0 gives a horizontal line, which has no slope triangles) and an intercept b. Without telling a partner the equation, give them three points on your line. The partner must (1) draw two slope triangles and show, with an angle argument, why they are similar; (2) compute m from each triangle and confirm the values agree; (3) derive the equation y = mx + b from the definition of slope, not by memory; and (4) check the equation at all three points. Then swap roles. Success criteria: the similarity argument names the right angles and the corresponding angles made with the horizontal legs; the derivation shows the step from (y − b) ÷ x = m to y = mx + b; every check is a substitution with the arithmetic written out.