Coordinate Geometry with Two Points: Line, Midpoint & Circle (Stewart 9e Q4)

Direct answer. With A(−7, 4) and B(5, −12): slope m = −4/3; line y = −(4/3)x − 16/3 with intercepts (−4, 0) and (0, −16/3); midpoint M(−1, −4); length AB = 20; perpendicular bisector y = (3/4)x − 13/4; and the circle with AB as diameter is (x + 1)² + (y + 4)² = 100.

Why this problem matters

Stewart bundles six sub-problems into Q4 because they form the spine of coordinate geometry. A student who can fluently chain slope → line → midpoint → distance → perpendicular bisector → circle has the toolkit for almost every two-dimensional geometry question in pre-calculus, AP Calc, SAT, SHSAT, and the New York State Regents. The reason it lives in the diagnostic — and the reason we cover it early at SOMATH — is that a student who is shaky on any one link in this chain will burn time recomputing it every time it shows up in a multi-step calculus problem later.

The other reason: every formula in this problem is symmetric in the two points. Once you write down two points cleanly on scratch paper, you can hand off the same labels (x₁, y₁) and (x₂, y₂) to slope, midpoint, and distance without re-labeling. Six answers, one labeling job.

(a) Slope of the line through A and B

The slope formula is the difference of y over the difference of x, in the same order:

m = (y₂ − y₁) / (x₂ − x₁) = (−12 − 4) / (5 − (−7)) = −16 / 12 = −4/3

The slope is negative, which matches the picture — point A is up and to the left, B is down and to the right, so the line falls as it moves right. The magnitude 4/3 says it drops 4 units for every 3 units of horizontal travel.

(b) Equation of the line and its intercepts

Use point-slope form anchored at A(−7, 4):

y − 4 = −(4/3)(x + 7)

Distribute and isolate y:

y = −(4/3)x − 28/3 + 4 = −(4/3)x − 28/3 + 12/3 = y = −(4/3)x − 16/3

To find the x-intercept, set y = 0: 0 = −(4/3)x − 16/3, so x = −4. The y-intercept is just the constant term, −16/3. Intercepts: (−4, 0) and (0, −16/3). Sanity check: plug B(5, −12) into the equation. −(4/3)(5) − 16/3 = −20/3 − 16/3 = −36/3 = −12. The line really does pass through B.

(c) Midpoint of segment AB

The midpoint formula averages the coordinates:

M = ((x₁ + x₂)/2, (y₁ + y₂)/2) = ((−7 + 5)/2, (4 + (−12))/2) = (−1, −4)

The midpoint is the geometric center of the two points and — critical for the circle problem at the end — it is the center of the circle whose diameter is AB. Keep M(−1, −4) visible on your scratch paper; you will need it twice more.

(d) Length of segment AB

Use the distance formula. Skip the unnecessary square-root-then-square-again by computing the squared distance first:

AB² = (x₂ − x₁)² + (y₂ − y₁)² = (5 − (−7))² + (−12 − 4)² = 12² + (−16)² = 144 + 256 = 400

Then AB = √400 = 20. Notice how clean: the differences are 12 and 16, the classic 3-4-5 triangle scaled by 4, with hypotenuse 5 · 4 = 20. Stewart picked these points so the arithmetic would be exact — that's a clue that a perfect square is hiding inside the distance.

(e) Perpendicular bisector of AB

The perpendicular bisector has two requirements: it passes through the midpoint, and it is perpendicular to the original line. We already have both pieces.

Midpoint M(−1, −4). Slope of AB is −4/3, so the perpendicular slope is the negative reciprocal, 3/4. Plug into point-slope:

y − (−4) = (3/4)(x − (−1)) ⇒ y + 4 = (3/4)(x + 1)

Distribute and isolate y:

y = (3/4)x + 3/4 − 4 = (3/4)x + 3/4 − 16/4 = y = (3/4)x − 13/4

Quick sanity check: does the midpoint satisfy this? Plug x = −1: (3/4)(−1) − 13/4 = −3/4 − 13/4 = −16/4 = −4. Yes — the line passes through (−1, −4), which is exactly what a bisector should do.

(f) Circle with AB as a diameter

This is the prize at the end of the chain. When the segment AB is the diameter of a circle, two facts fall out for free:

• The center of the circle is the midpoint of AB.
• The radius is half the length of AB.

We already computed both. Center M(−1, −4), and r = AB/2 = 20/2 = 10, so r² = 100. Plug into standard form (x − h)² + (y − k)² = r² with h = −1 and k = −4:

(x + 1)² + (y + 4)² = 100

The two plus signs inside the parentheses come from subtracting negative coordinates: (x − (−1)) = (x + 1) and (y − (−4)) = (y + 4). This is the same sign trap that ruins Q2 and Q3 — read every plus sign inside a squared binomial as "minus negative" and the center coordinates come out right.

The three mistakes worth memorizing

• Mixing up the order in the slope formula. The slope is (y₂ − y₁) / (x₂ − x₁), with the same point labeled "2" in both numerator and denominator. Students who swap one but not the other end up with the slope's sign flipped — which then flips the perpendicular slope, which then flips the bisector, which then flips the circle's geometry. Stack the two points vertically on scratch paper before subtracting and the order takes care of itself.
• Taking an unnecessary square root in the distance formula. Just like Q2, we want r², not r, for the circle equation. Compute the squared distance directly, plug it in, and only take a single square root at the very end if the problem asks for the actual length. Here AB² = 400 and we need (AB/2)² = 400/4 = 100 — a single division, no roots involved for the equation.
• Confusing the negative reciprocal with the negative. Perpendicular slopes multiply to −1. The perpendicular of slope −4/3 is +3/4, not +4/3 and not −3/4. Flip the fraction and flip the sign — both moves, in that order.

How SOMATH teaches this problem

At SOMATH we teach Stewart Q4 as one continuous chain, not six separate problems. Students label the two points once at the top of scratch paper. From there, slope feeds the line equation; the midpoint and slope feed the perpendicular bisector; the midpoint and distance feed the circle. Every formula reuses something we already wrote down. By the time students leave that lesson they have a worksheet that documents how a single labeled pair of points unlocks the whole of plane geometry — and they keep that habit for AP Calculus, where multi-step problems live or die on whether you reuse earlier work cleanly.

Our classroom at 226 W 79th St — Upper West Side, between Broadway and Amsterdam — runs small groups of three to five students with whiteboards on every wall. We don't hand out worksheet stacks. We introduce each technique with the why, drill it on three increasingly hostile problems, and tie it back to the calculus context where the student will meet it next. Q4 is the cornerstone problem we use to teach NYC independent-school students how to chain coordinate-geometry skills before they hit Stewart's calculus chapters.

FAQ

How do you find the slope between two points?

Use m = (y₂ − y₁) / (x₂ − x₁). Subtract the y-coordinates in the numerator and the x-coordinates in the denominator, in the same order. With A(−7, 4) and B(5, −12), slope is (−12 − 4)/(5 − (−7)) = −16/12 = −4/3.

What is the perpendicular bisector and how do you find its equation?

The perpendicular bisector of a segment passes through the segment's midpoint and is perpendicular to the segment. You need two ingredients: the midpoint (use the midpoint formula) and the perpendicular slope (negative reciprocal of the segment's slope). Plug both into point-slope form and simplify.

How is finding a circle from a diameter different from finding one from center and a point?

If you're given the diameter AB, the center is the midpoint of AB and the radius is AB/2. If you're given the center and a point on the circle (Stewart Q2), the center is handed to you and the radius is the distance from the center to that point. Same final form (x − h)² + (y − k)² = r² — different routes to (h, k) and r.

Why is the distance AB exactly 20?

The horizontal change is 12 and the vertical change is 16. That's a 3-4-5 right triangle scaled by 4, giving a hypotenuse of 5 · 4 = 20. Stewart deliberately picks numbers that produce perfect squares; if your distance comes out as an ugly radical, double-check your subtraction.

Will I see this exact problem type on the AP Calculus exam?

Not in this packaged form — AP Calc rarely asks pure coordinate-geometry questions. But the skills inside Q4 power optimization, parametric and polar problems, and any free-response question that involves distances between points on a curve. The diagnostic exists precisely so students don't have to relearn these tools mid-calculus.

Related posts

• Stewart Q2: Equation of a Circle from Center and a Point
• Stewart Q3: Center and Radius of a Circle in General Form
• Stewart Q1: Equation of a Line from Slope and a Point