Almost everything this course will draw — a point in space, a vector's tip, the corner of a region a system of equations carves out — starts as an ordered pair of numbers on a grid. You've plotted points and measured the distance between them since before you ever solved an equation with a matrix. This section isn't teaching that skill; it's re-sharpening it, and fixing the exact vocabulary and layout this course will keep coming back to: coordinates as an ordered pair, distance as a single number built from two differences, and the plane as the stage every later picture gets drawn on.
The coordinate plane, revisited
Think of the plane as a floor plan with two hallways crossing at a lobby. One hallway runs east–west, and the other runs north–south. The lobby where they cross is "home base," the origin. Giving someone a location always takes two numbers in a fixed order. "6 steps east, then 2 steps north" is a completely different spot from "2 steps east, then 6 steps north," even though the same two numbers are involved. That's all an ordered pair is: two directions, read in a fixed order, from a fixed home base. The four quadrants are just the four rooms that the crossing hallways create. Which room you're in is decided the instant you know which way each step went, east or west, north or south.
The Cartesian plane is built from two perpendicular number lines meeting at the origin, (0,0): a horizontal x-axis and a vertical y-axis. Every point in the plane corresponds to exactly one ordered pair (x, y) — ordered because (3, -5) and (-5, 3) name two different points, not the same one written twice. The x-coordinate is always read first and reports horizontal position (positive = right of the origin, negative = left); the y-coordinate is read second and reports vertical position (positive = above, negative = below). The two axes cut the plane into four quadrants, numbered I through IV counterclockwise starting from the upper right, where the signs of x and y are fixed within each: Quadrant I is (+,+), II is (-,+), III is (-,-), IV is (+,-). A point with a zero coordinate — (4, 0) or (0, -2) — sits on an axis, not inside any quadrant. One more term this section fixes before it's needed: distance is the single number that reports how far apart two points are, built from exactly the two coordinate differences the next box makes precise.
Legs and hypotenuse
Drag any vertex — legs a and b are the horizontal and vertical gap between two points, and hypotenuse c is the straight-line distance between them. Watch a^2+b^2=c^2 hold no matter how you drag.
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Check your understanding (counts toward your progress)
A student finds the distance between (3, 5) and (9, 1) as \sqrt{(3-9)^2+(5-1)^2} instead of \sqrt{(9-3)^2+(1-5)^2}. What happens to the answer?
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Squaring a difference erases its sign — try expanding both versions and compare.
A squared real number is never negative, and a square root returns the non-negative root, so distance can never be negative.
Subtracting in either order is perfectly valid — the formula only cares about the squared differences.
Why
(3-9)^2 = (-6)^2 = 36 and (9-3)^2 = 6^2 = 36 — same value either way. Swapping which point is 'first' flips the sign of each difference but not its square, so distance never changes.
Compare what happens when you square each difference -- squaring removes the sign, so check whether swapping the order changes anything once you square.
Find the distance between (-1, -1) and (2, 3), a segment that runs up and to the right.
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Why
Legs: Δx = 2-(-1) = 3, Δy = 3-(-1) = 4. Distance = sqrt(3^2 + 4^2) = sqrt(9+16) = sqrt(25) = 5.
Correct answer
Also accepted:
Why
Plug the two points into the distance formula \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} and simplify inside the square root first.
The ordered pairs (8, -3) and (-3, 8) describe:
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An ordered pair is ordered — swapping the x- and y-values moves you to a different location entirely, unless they happen to be equal.
Neither point has a zero coordinate, so neither one sits on an axis.
The origin is (0, 0); neither pair is that.
Why
Order matters in an ordered pair. (8, -3) is 8 right, 3 down; (-3, 8) is 3 left, 8 up — different locations.
Practice: The coordinate plane, one more look (4 problems · doesn’t affect your progress)
Which quadrant contains the point (-6, 1)? (For reference, the choices given are: Quadrant I; Quadrant II; Quadrant III; Quadrant IV.)
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Quadrant I needs both coordinates positive. Here x=-6 is negative.
Quadrant III needs both coordinates negative. Here y=1 is positive.
Quadrant IV needs x positive and y negative — this point has it backwards.
Why
x = -6 is negative (left of the origin) and y = 1 is positive (above the origin). Negative-x, positive-y is Quadrant II.
Quadrant II
The ordered pairs (8, -3) and (-3, 8) describe: (For reference, the choices given are: the same point; two different points; points on the same axis; the origin.)
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An ordered pair is ordered — swapping the x- and y-values moves you to a different location entirely, unless they happen to be equal.
Neither point has a zero coordinate, so neither one sits on an axis.
The origin is (0, 0); neither pair is that.
Why
Order matters in an ordered pair. (8, -3) is 8 right, 3 down; (-3, 8) is 3 left, 8 up — different locations.
two different points
Which of these points lies ON an axis rather than inside any quadrant?
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(3, -2) has both coordinates nonzero, so it sits inside Quadrant IV, not on an axis.
(-5, -5) has both coordinates nonzero, so it sits inside Quadrant III, not on an axis.
(1, 1) has both coordinates nonzero, so it sits inside Quadrant I, not on an axis.
Why
A point sits on an axis exactly when one of its coordinates is 0. (0, 7) has x=0, so it sits on the y-axis rather than inside any of the four quadrants.
A quadrant requires BOTH coordinates to be nonzero -- check each point for a zero coordinate first.
(0, 7)
A point in Quadrant II is reflected across the x-axis. Which quadrant does it land in?
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Reflecting across the x-axis flips the sign of y only, leaving x negative — Quadrant I needs a positive x, which never happens here.
Quadrant IV needs a positive x-coordinate; reflecting across the x-axis never changes the sign of x, so a negative x stays negative.
Reflecting across the x-axis flips the sign of y, so a point genuinely moves unless it started with y=0.
Why
Quadrant II is (-,+). Reflecting across the x-axis negates only the y-coordinate, giving (-,-) — Quadrant III.
Reflecting across the x-axis changes the sign of the y-coordinate only -- track what happens to the quadrant's sign pattern (-,+).
Quadrant III
The distance formula
Forget the formula for a second. Picture walking from one point to the other along the grid lines first: some steps east or west, then some steps north or south. That's an L-shaped path, not a straight one. Those two leg-lengths are exactly |x_2-x_1| and |y_2-y_1|. The direct, as-the-crow-flies distance is shorter than that L-shaped walk. The Pythagorean theorem is precisely the rule that turns those two step-counts into that straight-line distance. The midpoint is even simpler than it sounds. It's just the average location, found by averaging the two eastward positions and, separately, the two northward positions. That's the same kind of average you'd use to find the midpoint of a road trip from its start and end mile markers.
The distance between two points (x_1, y_1) and (x_2, y_2) is the length of the straight segment joining them. Build a right triangle from that segment: a horizontal leg of length |x_2 - x_1|, a vertical leg of length |y_2 - y_1|, and the segment itself as the hypotenuse. The Pythagorean theorem then gives the distance formula:
d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}.Squaring each difference erases the absolute value automatically — a negative difference squares to the same positive number a positive one would — so it never matters which point is labeled 1 and which is labeled 2; swapping the labels flips the sign of both differences but leaves both squares, and therefore d, unchanged. The midpoint of the same segment is the ordinary average of the two x-coordinates and the two y-coordinates separately:
M = \left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right).Worked example
Distance between two points
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1
Find the distance between (2, -3) and (7, 9).
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2these are the horizontal and vertical legs of the right triangle
Subtract coordinates in the same order for both pairs — here x_2-x_1 and y_2-y_1: x_2 - x_1 = 7 - 2 = 5, \qquad y_2 - y_1 = 9-(-3) = 12
these are the horizontal and vertical legs of the right triangle
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3
Square each leg and add: 5^2 + 12^2 = 25 + 144 = 169
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4
Take the square root — this is the hypotenuse, the straight-line distance: d = \sqrt{169} = \boxed{13}
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Worked example
Midpoint of a segment
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1
Find the midpoint of the segment joining (-4, 6) and (10, -2).
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2
Average the two x-coordinates: \frac{-4+10}{2} = \frac{6}{2} = 3
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3each coordinate is averaged on its own — never mix an x with a y
Average the two y-coordinates, independently of the x-average above: \frac{6+(-2)}{2} = \frac{4}{2} = 2
each coordinate is averaged on its own — never mix an x with a y
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4
Combine the two averages into one ordered pair: M = \boxed{(3,\ 2)}
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Check your understanding (counts toward your progress)
Find the distance between (4, 6) and (-2, -2), a segment that runs down and to the left.
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Why
Legs: Δx = 4-(-2) = 6, Δy = 6-(-2) = 8. Distance = sqrt(6^2 + 8^2) = sqrt(36+64) = sqrt(100) = 10.
Correct answer
Also accepted:
Why
Find the horizontal and vertical differences between the points before squaring and adding them.
Find the distance between (1, 2) and (1, 9).
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Why
Same x-coordinate, so this is a purely vertical segment: the distance is just the difference in y, |9 - 2| = 7 (the general formula agrees: sqrt(0^2 + 7^2) = 7).
Correct answer
Also accepted:
Why
Points A(0, 0), B(6, 0), and C(6, 8) form a triangle. Find the length of side AC (from A straight to C, not along the other two sides).
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Why
AC's legs are Δx = 6-0 = 6 and Δy = 8-0 = 8, so AC = sqrt(6^2+8^2) = sqrt(36+64) = sqrt(100) = 10 — note this is longer than AB + BC would suggest going 'around', since AC cuts straight across.
Correct answer
Also accepted:
Why
Apply the distance formula directly to points A and C -- you don't need B at all for this particular side.
A segment has midpoint (4, -1) and one endpoint (1, 5). Find the x-coordinate of the other endpoint.
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Why
Midpoint x-formula: 4 = (1 + x2)/2, so 8 = 1 + x2, so x2 = 7 (the midpoint is exactly as far from each endpoint, so 4 is 3 past 1, meaning the other endpoint is 3 past 4, at 7).
Correct answer
Also accepted:
Why
The midpoint's coordinate is the average of the two endpoints' coordinates -- set up that average equation and solve for the unknown endpoint.
Practice: Distance and midpoint, start to finish (5 problems · doesn’t affect your progress)
Find the distance between (1, 2) and (1, 9), a purely vertical segment.
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Why
Same x-coordinate, so this is a purely vertical segment: the distance is just the difference in y, |9 - 2| = 7 (the general formula agrees: sqrt(0^2 + 7^2) = 7).
Correct answer
Also accepted:
Why
Find the distance between (-1, -1) and (2, 3), moving right and up.
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Why
Legs: Δx = 2-(-1) = 3, Δy = 3-(-1) = 4. Distance = sqrt(3^2 + 4^2) = sqrt(9+16) = sqrt(25) = 5.
Correct answer
Also accepted:
Why
Plug the two points into the distance formula \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} and simplify inside the square root first.
The midpoint of the segment from (0, 0) to (6, 10) is: (For reference, the choices given are: (6, 10); (3, 5); (3, 10); (6, 5).)
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That's the endpoint, not the midpoint — average the two x-values and the two y-values separately.
The x-coordinate wasn't averaged — (0+6)/2 = 3, not 6.
The y-coordinate wasn't averaged — (0+10)/2 = 5, not 10.
Why
Midpoint = ((0+6)/2, (0+10)/2) = (3, 5) — the average x and the average y, found independently of each other.
(3, 5)
A segment has midpoint (4, -1) and one endpoint (1, 5). Find the x-coordinate of the other endpoint. (Work through it specifically for (4, -1) and (1, 5) and x.)
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Why
Midpoint x-formula: 4 = (1 + x2)/2, so 8 = 1 + x2, so x2 = 7 (the midpoint is exactly as far from each endpoint, so 4 is 3 past 1, meaning the other endpoint is 3 past 4, at 7).
Correct answer
Also accepted:
Why
The midpoint's coordinate is the average of the two endpoints' coordinates -- set up that average equation and solve for the unknown endpoint.
For which pair of points does the distance formula reduce to just |y_2 - y_1|, with no square root needed? (For reference, the choices given are: (3, 2) and (8, 2); (3, 2) and (3, 9); (3, 2) and (8, 9); (0, 0) and (5, 5).)
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Here x is fixed instead — that makes it a horizontal segment, reducing to |x2 - x1|, not |y2 - y1|.
Neither coordinate matches here, so both legs are nonzero and the full formula is needed.
Neither coordinate matches here either — both legs are nonzero (both equal to 5, in fact).
Why
(3,2) and (3,9) share the same x-coordinate, so Δx = 0. The formula sqrt(0^2+(y2-y1)^2) simplifies to just |y2-y1|, since the horizontal leg vanishes entirely and only the vertical leg remains.
The distance formula simplifies to just |y_2-y_1| exactly when the x-coordinates of the two points are equal, so the (x_2-x_1)^2 term drops out.
(3, 2) and (3, 9)
Where this shows up: setting the stage for vectors
Where this goes next
Linear Algebra
Every idea in this section is about to get a second life. Starting next chapter, an ordered pair (x,y) is also a vector — the same two numbers, now read as an arrow from the origin instead of a location — and the distance formula becomes the length of that arrow, \sqrt{x^2+y^2}. The difference-of-coordinates trick used to find the distance between two points becomes vector subtraction: the arrow from one point to another is (x_2-x_1,\, y_2-y_1). Even the midpoint formula survives unchanged, as the average of two position vectors. None of the arithmetic here is being thrown away once the course starts talking about vectors and matrices — it's the exact computation, wearing a new name and a new picture.
What's next
With coordinates, distance, and the midpoint fixed as working vocabulary, the next piece of the toolkit is angle: how to measure the angle between two directions and what cosine has to do with it — the last ingredient this readiness chapter needs before sigma notation, and eventually vectors themselves, can be introduced.