Math

Coordinate Geometry

6 min readMedium5-question drill

Coordinate geometry questions test five things: distance, midpoint, slope, line equations, and parallel/perpendicular relationships. Master those five formulas and you can solve almost every coordinate-grid problem on the SAT.

Where this lesson fits

Before

Linear Functions→Right Triangles→

Coordinate Geometry

You are here

→Circles →Systems of Equations

The five core formulas:

1. Distance between two points:

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

This is just the Pythagorean theorem applied to coordinate differences.

2. Midpoint of two points:

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

The average of the $x$ 's and the average of the $y$ 's.

3. Slope between two points:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}$

4. Line equation forms:

Slope-intercept: $y = mx + b$ . $m$ is slope; $b$ is the $y$ -intercept (where the line crosses the $y$ -axis).
Point-slope: $y - y_1 = m(x - x_1)$ . Useful when you know a point and slope.
Standard: $Ax + By = C$ .

5. Parallel and perpendicular slopes:

Parallel lines: same slope. $m_1 = m_2$ .
Perpendicular lines: slopes are negative reciprocals. $m_1 \cdot m_2 = -1$ .

Slope and line relationships at a glance

Original slope	Parallel slope	Perpendicular slope
$2$	$2$	$-\frac{1}{2}$
$-3$	$-3$	$\frac{1}{3}$
$\frac{2}{5}$	$\frac{2}{5}$	$-\frac{5}{2}$
$-\frac{3}{4}$	$-\frac{3}{4}$	$\frac{4}{3}$
$0$ (horizontal)	$0$	undefined (vertical)

SAT-typical setups:

"Find the distance from $(2, 3)$ to $(7, 15)$ ." → $\sqrt{(7-2)^2 + (15-3)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ .
"Find the line through $(2, 3)$ with slope $-1/2$ ." → $y = -\frac{1}{2}x + 4$ (using point-slope and simplifying).
"Find the line perpendicular to $y = 2x + 1$ passing through $(4, 5)$ ." → perpendicular slope = $-1/2$ . Use point-slope: $y - 5 = -\frac{1}{2}(x - 4) \Rightarrow y = -\frac{1}{2}x + 7$ .

Three forms of a line equation

Form	Equation	When to use
Slope-intercept	$y = mx + b$	When you know slope and y-intercept (or want them visible)
Point-slope	$y - y_1 = m(x - x_1)$	When you know a point and the slope
Standard	$Ax + By = C$	When you need to compare or solve a system

Common SAT formulas in disguise:

Equation of a circle — also coordinate geometry: $(x-h)^2 + (y-k)^2 = r^2$ . Center $(h, k)$ , radius $r$ .
Distance from point to line — usually solved by finding perpendicular and computing distance.

Sign trap: parallel slopes match exactly; perpendicular slopes flip sign AND reciprocate. So:

Slope 3 → perpendicular slope $-1/3$ .
Slope $-2$ → perpendicular slope $1/2$ .
Slope $\frac{2}{5}$ → perpendicular slope $-\frac{5}{2}$ .

Special slopes:

Horizontal line: slope = 0. Equation: $y =$ constant.
Vertical line: slope = undefined. Equation: $x =$ constant.
A horizontal and vertical line are perpendicular even though their slopes don't multiply to $-1$ (the rule breaks down because vertical's slope is undefined).

Quick check

Identify which formula to use (distance, midpoint, slope, line equation, parallel/perpendicular). Most coordinate geometry problems are one or two of these in combination.

What is the distance between (0, 0) and (3, 4)?

Worked examples

Example 1

Line $\ell$ has equation $y = 3x + 2$ . What is the slope of a line perpendicular to $\ell$ ?

Example 2

What is the equation of the line passing through $(1, 4)$ and $(5, 12)$ ?

Common pitfalls

Forgetting the negative on perpendicular slopes

Perpendicular slopes are NEGATIVE reciprocals. Slope 3 → perpendicular is $-\frac{1}{3}$ , not $\frac{1}{3}$ . Always flip the sign in addition to taking the reciprocal.

Mixing up rise/run direction

Slope = $\frac{y_2 - y_1}{x_2 - x_1}$ . Subtract $y$ 's on top and $x$ 's on bottom in the SAME order. If you flip one but not the other, you'll get the wrong sign.

Wrong formula for midpoint

Midpoint AVERAGES the coordinates: $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ . Don't subtract — that would give half the distance vector, not the midpoint.

Forgetting to square in distance formula

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ . The squares are inside the square root. Forgetting them gives you $|x_2 - x_1| + |y_2 - y_1|$ — the taxicab distance, not the straight-line distance.

Key takeaways

Distance: $\sqrt{(\Delta x)^2 + (\Delta y)^2}$ . Midpoint: average each coordinate.
Slope: $\frac{y_2 - y_1}{x_2 - x_1}$ . Subtract $y$ 's and $x$ 's in the same order.
Slope-intercept form: $y = mx + b$ . Point-slope: $y - y_1 = m(x - x_1)$ .
Parallel slopes are equal. Perpendicular slopes are NEGATIVE reciprocals.
Horizontal: $y = c$ (slope 0). Vertical: $x = c$ (slope undefined).

Watch & learn

Curated Khan Academy walkthroughs on Coordinate Geometry. They're complementary to this lesson — watch one if a written explanation isn't clicking, or after to reinforce.

Try it yourself

5 practice questions on Coordinate Geometry, drawn from the question bank. The tutor is one click away if you get stuck.

Where this lesson fits

Before

Linear Functions→Right Triangles→

Coordinate Geometry

You are here

→Circles →Systems of Equations