Coordinate geometry: distance, midpoint and shapes on the grid
Two subtractions and a square root do most of it — the marks are lost on the signs.
What are the coordinates of the point?
In the xy-plane, point M(−1, 4) is the midpoint of
segment AB. If point A has coordinates (5, −2), what
are the coordinates of point B?
A midpoint is an average, so undo it: B = 2M − A. That gives x = 2(−1) − 5 = −7 and y = 2(4) − (−2) = 10. B reports the step from A to M, (−6, 6), instead of the point you land on. C averages a second time, returning the midpoint of A and M. D swaps the roles and computes 2A − M.
1
The distance formula is the Pythagorean theorem wearing coordinates.
Two points on a grid are the ends of a hypotenuse. The legs run along the gridlines, so their lengths are just subtractions.
across = 9 − (−3) = 12
the x-difference — minus a minus adds
up = 7 − 2 = 5
the y-difference
d = √12² + 5² = √169 = 13
Pythagoras on the two legs
The drawing is to scale: the horizontal leg spans 12 grid squares and the vertical
leg spans 5, so the slanted side really is the 13 of a 5-12-13 triangle. Sketching
the two legs is the safest way to keep the signs straight.
2
Written as a formula, it is the same two subtractions.
d = √ (x₂ − x₁)² + (y₂ − y₁)²
a² + b² = c²
the theorem
legs, then hypotenuse
the same thing
a and b are the two differences
c is the distance
Square first, add second, root last
√144 + 25 is √169 = 13. It is not√144 + √25 = 17 — a square
root never splits across a plus. And the order of the points cannot matter:
(−3 − 9)² = (−12)² = 144, exactly the same as (9 − (−3))². Squaring destroys the
sign, which is why distance has no direction and slope does.
Let Desmos do the arithmetic
The Digital SAT keeps the graphing calculator open the whole time. Type the two
points as (−3, 2) and (9, 7) to see them plotted, or type
sqrt((9−(−3))^2 + (7−2)^2) straight into a line and read 13. It is the fastest
way to catch a dropped negative.
3
The midpoint is the average of the coordinates — nothing more.
M = (
x1 + x22
,
y1 + y22
)
x: −3 + 92 = 3
add the two x's, halve
y: 2 + 72 = 4.5
add the two y's, halve
M = (3, 4.5)
the midpoint of A(−3, 2) and B(9, 7)
Add for a midpoint, subtract for a distance. A midpoint of a segment with a
negative endpoint can still be positive, and a coordinate ending in .5 is perfectly
normal — do not round it off.
The midpoint is a point, not a number
It has an x and a y, and the two never mix: the x's only ever average with x's.
Keeping them in two columns is what stops −5 + 72 from picking up a stray 2
from the other coordinate.
4
Given the midpoint and one end, undo the average.
This is the SAT's usual version — and the one where the signs bite. The opening problem: M(−1, 4) is the midpoint of AB and A is (5, −2).
Point
x
y
endpoint A
5
−2
midpoint M
−1
4
endpoint B
−7
10
−1 = 5 + x2
the x equation
−2 = 5 + x, so x = −7
× 2 both sides, then − 5
4 = −2 + y2
the y equation
8 = −2 + y, so y = 10
× 2 both sides, then + 2
B = (−7, 10)
The one-line shortcut
missing end = 2 × midpoint − known end
2(−1) − 5 = −7 · 2(4) − (−2) = 10
Or count the step, twice
Going from A(5, −2) to M(−1, 4) is 6 left and 6 up. Take exactly that step again
from M and you land on (−7, 10). Stopping after the first step, and answering
(−6, 6), is reporting the step instead of the point — that is choice B on the
opening problem.
5
To name a shape from its vertices, use slopes for angles and distances for sides.
Every "which of the following is quadrilateral ABCD" question comes down to these two tests.
A horizontal side has slope 0 and a vertical side has no slope at all, so the
"multiply to −1" test cannot be run on that pair — but horizontal and vertical are
obviously perpendicular, so read it off and move on.
6
Worked: a quadrilateral that turns out to be a square.
Quadrilateral PQRS has vertices P(1, 1), Q(5, 2), R(4, 6) and S(0, 5). What kind of quadrilateral is it?
slope PQ = 2 − 15 − 1 = 14
and slope SR is 6 − 54 − 0 = 14 too — parallel
slope QR = 6 − 24 − 5 = −4
and slope PS is 5 − 10 − 1 = −4 too — parallel
14 × (−4) = −1
so ∠Q is a right angle — it is a rectangle
PQ = √4² + 1² = √17
4 across, 1 up
QR = √(−1)² + 4² = √17
adjacent sides equal — so it is a square
PQRS is a square of side √17, so its area is 17
Tilted does not mean irregular
Nothing here is horizontal or vertical, and every side length is irrational, yet
the figure is a perfect square. That is exactly why the SAT tests it with slopes
and distances rather than with a picture — and why √17 × √17 = 17 gives the area
without ever evaluating the root.
✓
Check: put the answer back into the question.
midpoint5 + (−7)2 = −1and −2 + 102 = 4 ✓
distance144 + 25 = 169√169 = 13, and 13 > 12 ✓
square(14)(−4) = −1right angle, and both sides √17 ✓
sensible?the hypotenuse is the longest sidealways, on any grid ✓
Remember the flow
Subtract for distance · Average for midpoint · Double and subtract for the missing end
Slopes decide the angles; distances decide the sides.
Now you try
Practice problem
What is the value of the length?
In the xy-plane, point M(2, 1) is the midpoint of
segment AB, and point A has coordinates (−1, 5).
What is the length of segment AB?
B = 2M − A = (2(2) − (−1), 2(1) − 5) = (5, −3), so AB = √6² + 8² = √100 = 10. Faster still: AM = √3² + 4² = 5 and AB is twice that. A stops at AM, half the segment — the slip the problem is built around. C adds the legs, 6 + 8, instead of using the theorem. D forgets the square root and reports 100.
Hint: you do not have to find B at all. The midpoint splits AB into two equal
halves, so measure A to M and double it — and check at the end which length the
question actually asked for.
