Similar Triangles Explain Constant Slope

A slope triangle for two points on a slanted line (neither horizontal nor vertical) is a right triangle whose third side is the piece of the line between the two points. Its legs, the two sides that meet at the right angle, are one horizontal and one vertical. Its run is the length of the horizontal leg and its rise is the length of the vertical leg, both in the same unit. Moving from the left point to the right point, $x$ grows by the run. On a line that goes up from left to right, $y$ grows by the rise, so the slope is rise ÷ run.

In the figure below, triangle $ABC$ is a slope triangle for points $A$ and $B$. The square mark at $C$ is a given: angle $ACB$ is a right angle. The run is $CA = 3$ units and the rise is $CB = 2$ units. These lengths are stated here, not measured from the drawing. So the slope from $A$ to $B$ is $\frac{2}{3}$.

The figure below keeps triangle $ABC$, with its right angle at $C$, $CA = 3$ and $CB = 2$, and adds a second slope triangle, $DEF$, for points $D$ and $E$ farther up the same line. The square mark at $F$ gives angle $DFE$ as a right angle. The run of triangle $DEF$ is $FD = 6$ units and its rise is $FE = 4$ units, so the slope from $D$ to $E$ is $\frac{4}{6} = \frac{2}{3}$. That matches the slope from $A$ to $B$, even though triangle $DEF$ is bigger.

The match is no accident. Two triangles are similar when one is a scaled copy of the other: matching angles are equal, and every side length is multiplied by one scale factor. If two angles of one triangle equal two angles of another, the triangles are similar. Check two pairs of angles:

• Right angles. The square marks give angle $ACB$ = angle $DFE$ = $90^\circ$.
• Angles with the line. Both runs are horizontal, so $CA$ and $FD$ lie on parallel lines, and the slanted line crosses both. That makes angle $CAB$ and angle $FDE$ corresponding angles (angles in matching positions at the two crossings), and a line crossing parallel lines makes corresponding angles equal.

So triangle $ACB$ is similar to triangle $DFE$. Matching sides join matching angles, so run $CA$ matches run $FD$, and rise $CB$ matches rise $FE$. The scale factor is $FD \div CA = 6 \div 3 = 2$, so $FD = 2 \cdot 3$ and $FE = 2 \cdot 2$. The nonzero factor $2$ cancels in rise ÷ run:

$\frac{FE}{FD} = \frac{2 \cdot 2}{2 \cdot 3} = \frac{2}{3} = \frac{CB}{CA}$

The coordinate plane below shows the same line, $y = \frac{2}{3}x + 1$, with the $x$-axis horizontal and the $y$-axis vertical. Each point sits at exact coordinates: $A(0, 1)$, $C(3, 1)$, $B(3, 3)$, $D(6, 5)$, $F(12, 5)$ and $E(12, 9)$. Each right-angle vertex, $C$ or $F$, has the same $y$-coordinate as the point to its left and the same $x$-coordinate as the point above it, so every run is horizontal and every rise is vertical. Here each run is a change in $x$ and each rise is a change in $y$: $CA = 3 - 0 = 3$, $CB = 3 - 1 = 2$, $FD = 12 - 6 = 6$ and $FE = 9 - 5 = 4$, the values given before. The graph of a proportional relationship always passes through the origin, but this line crosses the $y$-axis at $(0, 1)$, so it is not proportional. The similar-triangle argument never used the origin, so the slope is still $\frac{2}{3}$ for every pair of points.

Takeaway: On a line that is neither horizontal nor vertical, any two slope triangles each have a right angle and have equal angles with the line, so they are similar. Similarity multiplies the rise and the run by the same positive factor, which cancels in rise ÷ run, so every pair of points on the line gives the same slope. Next, you'll find and correct errors in calculating slope.