Coordinate Geometry: SAT Math Lesson

Use distance, midpoint, and slope to read geometry off coordinates: classify figures, test points against circles, and connect lines, distances, and circle equations.

Geometry on the grid

Place a figure on the grid and its shape becomes arithmetic. Distance, midpoint, and slope are the three instruments that read geometry straight off the coordinates.

This topic is where two halves of the Geometry and Trigonometry domain meet. Coordinate geometry sets a figure down on the grid you have used since the Algebra domain, and at that moment its geometry (lengths, midpoints, the angles between its sides) turns into something you can compute from the points alone. You no longer measure; you calculate. Three instruments do nearly all of it: distance gives a length, midpoint gives a center, and slope gives a direction. Master those three and you can take a heap of coordinates and read off, with certainty, whether a figure is a parallelogram, a right triangle, or a circle of a given radius.

The distance formula and the Pythagorean theorem

The distance between two points is the Pythagorean theorem in disguise, and seeing why makes the formula unforgettable. Drop a horizontal segment and a vertical segment between the two points and you have built a right triangle: the horizontal leg is the difference in the $x$ -coordinates, the vertical leg the difference in the $y$ -coordinates, and the distance you want is the hypotenuse. So the theorem gives it directly.

$A coordinate plane with a segment from (1, 1) to (7, 5) as the hypotenuse of a right triangle with horizontal leg 6 and vertical leg 4, labeled d equals the square root of 6 squared plus 4 squared$

Figure 1. The distance between two points is the hypotenuse of a right triangle whose legs are the coordinate differences.

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Worked: compute a distance. Find the distance between $(1, 1)$ and $(7, 5)$ . The legs are the coordinate differences, $7 - 1 = 6$ and $5 - 1 = 4$ : $d = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$ . It is just the Pythagorean theorem with the legs handed to you by subtraction.

Watch for Pythagorean triples hiding in the differences. Differences of $6$ and $8$ give $10$ ; differences of $5$ and $12$ give $13$ . Spotting the triple saves the arithmetic. Squaring also removes the sign, so it does not matter which point you call first, as long as you subtract in the same order for both coordinates.

The midpoint: average the coordinates

The midpoint, the point exactly halfway between two others, is even simpler, because halfway in two dimensions is just halfway in each one separately. So you average the two $x$ -coordinates and average the two $y$ -coordinates. That is the entire idea.

$\text{midpoint} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

Worked: find a midpoint, and reverse it. The midpoint of $(1, 1)$ and $(7, 5)$ averages each coordinate: $\left(\frac{1 + 7}{2}, \frac{1 + 5}{2}\right) = (4, 3)$ . The same idea runs backward. If the midpoint of a segment is $(3, 5)$ and one endpoint is $(-1, 2)$ , the other endpoint is as far past the midpoint as the first falls short, so double the midpoint and subtract the known endpoint: $(2 \cdot 3 - (-1),\ 2 \cdot 5 - 2) = (7, 8)$ . Averaging finds the center; doubling-and-subtracting recovers an endpoint.

The reverse move matters for circles too: the center of a circle is the midpoint of any diameter, so a diameter's two endpoints give you the center by averaging, and the radius by the distance from that center to either endpoint.

Slope for parallel and perpendicular relationships

Slope, met first in the Algebra domain, becomes a geometric instrument here: it measures the direction a segment points, as rise over run, $\frac{y_2 - y_1}{x_2 - x_1}$ . Two facts about direction unlock most coordinate figures. Parallel lines point the same way, so they have equal slopes. For two nonvertical, nonhorizontal perpendicular lines, the slopes are negative reciprocals and their product is $-1$ . A vertical line is perpendicular to a horizontal line, which is the edge case where the product rule does not apply. A slope of $2$ is perpendicular to a slope of $-\frac12$ ; a slope of $\frac34$ to a slope of $-\frac43$ .

$Two perpendicular lines through the origin on a coordinate plane, one with slope 2 and one with slope negative one half, with a right angle marked and the note that slopes multiply to negative 1$

Figure 2. Parallel lines share a slope; perpendicular lines have slopes that are negative reciprocals, multiplying to $-1$ .

To build a negative reciprocal, do two things: flip the fraction, then change the sign. For $-\frac25$ , flipping gives $-\frac52$ and changing the sign gives $\frac52$ . Always check by multiplying: $-\frac25 \cdot \frac52 = -1$ .

The edge cases come up often. A line through $(4, -2)$ and $(4, 6)$ has the same $x$ -coordinate throughout, so it is vertical with undefined slope; any line perpendicular to it is horizontal, with slope $0$ . A line through $(2, 7)$ and $(9, 7)$ is horizontal with slope $0$ ; a perpendicular line is vertical, with undefined slope. The negative-reciprocal rule cannot be applied here, since $0$ has no reciprocal and an undefined slope cannot be inverted.

Use coordinates to recover geometric information

Now the three instruments combine, and this is where the test concentrates its harder questions. Given the vertices of a figure, you classify it by computing the right quantities: slopes tell you which sides are parallel (a parallelogram) or perpendicular (a right angle), and distances tell you which sides are equal (an isosceles triangle, a rhombus, a square). The strategy is to decide what you need to prove and compute exactly that: equal slopes for parallel sides, a slope product of $-1$ for a right angle, equal distances for equal sides.

$A quadrilateral ABCD plotted on a coordinate plane with vertices A at lower left, B at lower right, C at upper right, and D at upper left, marked as a parallelogram because AB is parallel to DC and AD is parallel to BC$

Figure 3. Computing the four side slopes shows opposite sides are parallel, so this quadrilateral is a parallelogram.

Worked: classify a figure. A quadrilateral has vertices $A(1, 1)$ , $B(5, 2)$ , $C(6, 5)$ , $D(2, 4)$ . Test for a parallelogram by comparing opposite sides' slopes. Slope of $AB$ is $\frac{2 - 1}{5 - 1} = \frac14$ ; slope of $DC$ is $\frac{5 - 4}{6 - 2} = \frac14$ : equal, so $AB \parallel DC$ . Slope of $AD$ is $\frac{4 - 1}{2 - 1} = 3$ ; slope of $BC$ is $\frac{5 - 2}{6 - 5} = 3$ : equal, so $AD \parallel BC$ . Both pairs of opposite sides are parallel, so the figure is a parallelogram. To check further for a rectangle, you would test whether adjacent slopes multiply to $-1$ .

A shortcut for a missing parallelogram vertex: the diagonals of a parallelogram share a midpoint. So if $A$ , $B$ , $C$ are three vertices in order, the fourth vertex is $D = A + C - B$ , found by adding the opposite vertices' coordinates and subtracting the remaining one. For the figure above, $D = (1 + 6 - 5,\ 1 + 5 - 2) = (2, 4)$ .

For a triangle, choose the test that matches the claim. To show a right angle, find the slopes of the two sides at that vertex and check that their product is $-1$ (or compare squared distances with the Pythagorean theorem). To show a triangle is isosceles, compute two side lengths and compare. Area is often easiest from a base and height: if a side runs along an axis or is horizontal or vertical, its length is just a difference of coordinates, and the height is a difference of the other coordinate.

Connect line equations, distances, and circle equations

Coordinate geometry also closes the loop with the equations you already know. A line through two plotted points is found by computing its slope and reading an intercept; a circle is the set of points a fixed distance, its radius, from a center, which is why $(x - h)^2 + (y - k)^2 = r^2$ is just the distance formula set equal to the radius. That single view answers a family of questions: a point lies on a circle when its distance from the center equals the radius, inside when the distance is less, and outside when it is greater. The instruments of this topic and the equations of the earlier ones are the same ideas, seen from two sides.

Worked: test a point against a circle. Does $(6, 8)$ lie on the circle $x^2 + y^2 = 100$ , centered at the origin with radius $10$ ? Find the point's distance from the center: $\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ . The distance equals the radius, so the point lies exactly on the circle, a $6$ - $8$ - $10$ right triangle hiding in the coordinate plane.

Two more moves round out the toolkit. To find where a circle crosses an axis, set the other variable to $0$ and solve; for $(x - 3)^2 + (y - 4)^2 = 25$ at $y = 0$ , you get $(x - 3)^2 = 9$ , so $x = 0$ or $x = 6$ . And if a circle is tangent to the $x$ -axis, its radius equals the absolute value of the $y$ -coordinate of its center; if it is tangent to the $y$ -axis, its radius equals the absolute value of the $x$ -coordinate of its center.

Summary

On the coordinate plane, geometry becomes arithmetic through three instruments: distance, the Pythagorean hypotenuse of the coordinate differences; midpoint, the average of the coordinates; and slope, the direction, with parallel lines sharing it and perpendicular lines turning it into its negative reciprocal. Combine them to classify any figure (equal slopes for parallel sides, negative-reciprocal slopes for nonvertical, nonhorizontal perpendicular lines, with vertical and horizontal lines handled as the special case, equal distances for equal sides) and recall that a circle's equation is simply the distance formula set to its radius. Decide what you must show, compute exactly that, and the coordinates tell you the shape.

Coordinate Geometry