Slope using coordinates

From two points to a number. Let a line pass through two distinct points $P(x_1, y_1)$ and $Q(x_2, y_2)$, with $x_1 \neq x_2$. Its slope is defined as the change in $y$ divided by the change in $x$ between the two points:

where: $m$ is the slope of line $PQ$, $(x_1,y_1)$ and $(x_2,y_2)$ are the two given points, and the difference $x_2-x_1$ must be nonzero for the ratio to be defined.

It agrees with the angle definition. $(y_2-y_1)$ is exactly the vertical rise between the points and $(x_2-x_1)$ is exactly the horizontal run, so their ratio equals $\tan\theta$, where $\theta$ is the line's inclination — this coordinate formula is not a new idea, just a computable version of $m=\tan\theta$.

Order doesn't matter, as long as it's consistent. Swapping which point you call "first" and which "second" flips the sign of both the numerator and the denominator, so the ratio is unchanged: $\dfrac{y_2-y_1}{x_2-x_1} = \dfrac{y_1-y_2}{x_1-x_2}$. What you must never do is swap only one of the two subtractions.

Two special cases to recognise on sight. If $x_1 = x_2$ the line is vertical and the denominator is zero, so the slope is undefined — not zero, not infinite in a usable sense, simply not defined as a number. If $y_1 = y_2$ the line is horizontal and the numerator is zero, so the slope is exactly $0$.

Why the ratio never depends on which two points you pick. Take any two points on a line and drop a horizontal segment from the left one and a vertical segment from the right one; they meet at a right angle, forming a right triangle whose legs have lengths $x_2-x_1$ and $y_2-y_1$. The angle between the line and its horizontal leg is the inclination $\theta$ itself, so $\tan\theta = \dfrac{y_2-y_1}{x_2-x_1}$. Any other pair of points on the same line forms a similar right triangle — same angle $\theta$, proportionally scaled legs — so the ratio of vertical to horizontal leg is identical every time. That is precisely why slope is a property of the whole line, not of any one pair of points on it.

Problem. Find the slope of the line through $A(2,-3)$ and $B(6,5)$, then use the same method to check whether $A(1,1)$, $B(3,5)$, and $C(5,9)$ are collinear.

Step 1. Apply the formula to $A(2,-3)$ and $B(6,5)$ with $(x_1,y_1)=(2,-3)$ and $(x_2,y_2)=(6,5)$:

Step 2. Sanity-check the sign: as $x$ increases from $2$ to $6$, $y$ increases from $-3$ to $5$, so the line rises left to right — a positive slope of $2$ is consistent with the picture.

Step 3. Now take the second, unrelated set of points $A(1,1)$, $B(3,5)$, $C(5,9)$, and find the slope of $AB$: $m_{AB} = \dfrac{5-1}{3-1} = \dfrac{4}{2} = 2$. Find the slope of $BC$: $m_{BC} = \dfrac{9-5}{5-3} = \dfrac{4}{2} = 2$.

Step 4. Since $m_{AB} = m_{BC} = 2$ and both were computed through the shared point $B$, the three points lie on one straight line. As a cross-check, the slope of $AC$ directly is $m_{AC} = \dfrac{9-1}{5-1} = \dfrac{8}{4} = 2$, agreeing with both.

Final answer: the slope of $AB$ is $\mathbf{m=2}$, and $A(1,1)$, $B(3,5)$, $C(5,9)$ are collinear because $m_{AB} = m_{BC} = m_{AC} = 2$.

Competitive stretch

Find $k$ so that $A(1,4)$, $B(3,k)$, $C(7,2)$ are collinear. Set $m_{AB} = m_{AC}$: $\dfrac{k-4}{3-1} = \dfrac{2-4}{7-1}$, so $\dfrac{k-4}{2} = \dfrac{-2}{6} = -\dfrac{1}{3}$, giving $k-4 = -\dfrac{2}{3}$, so $k = \dfrac{10}{3}$. This kind of "find the unknown coordinate that keeps points collinear" question is a favourite in competitive papers precisely because it needs nothing beyond the slope formula, applied twice.

• Mixing up which coordinate goes on top: it is always $\dfrac{y_2-y_1}{x_2-x_1}$, change in $y$ over change in $x$, never the reverse.
• Subtracting in a criss-cross, mismatched order — for example computing $\dfrac{y_2-y_1}{x_1-x_2}$ — which flips the sign of the slope and gives a wrong answer even though the arithmetic "looks" fine.
• Forgetting to check $x_1 \neq x_2$ before dividing; if the two points share an x-coordinate the line is vertical and the slope is undefined, not zero.
• In collinearity checks, comparing the wrong pair of slopes, or computing them with inconsistent point order, and getting a spurious mismatch that suggests three points are not collinear when they actually are.

Ask the tutor: want a quick set of coordinate-slope and collinearity practice problems, or a walk-through of how this formula feeds into the point-slope form of a line's equation?