4.4 Coordinate Geometry

ALEKS Math Placement Study Guide Exam

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1 Chapter 1: ALEKS PPL Overview & Strategy

1.1 What ALEKS PPL Is 1.2 How Placement Works 1.3 The Prep & Learning Module & Retake Strategy

2 Chapter 2: Arithmetic & Number Sense

2.1 Real Numbers & Operations 2.2 Fractions, Decimals & Percents 2.3 Ratios, Proportions & Unit Conversion

3 Chapter 3: Algebra Essentials

3.1 Linear Equations & Inequalities 3.2 Exponents & Polynomials 3.3 Factoring & Rational Expressions 3.4 Quadratic Equations

4 Chapter 4: Functions, Graphs & Geometry

4.1 Lines, Slope & Graphing 4.2 Functions & Function Notation 4.3 Geometry & Measurement 4.4 Coordinate Geometry

5 Chapter 5: Advanced Readiness — Precalculus & Trig

5.1 Exponential & Logarithmic Expressions 5.2 Radicals, Rational Exponents & Complex Numbers 5.3 Trigonometry Basics 5.4 Sequences, Functions Review & Exam Readiness

4.4 Coordinate Geometry

Key Takeaways Explain This Topic AI

Distance formula: d = √((x₂ − x₁)² + (y₂ − y₁)²) — it is the Pythagorean theorem on the plane.

Midpoint formula: M = ((x₁ + x₂)/2, (y₁ + y₂)/2) — average the x's and the y's.

A circle centered at (h, k) with radius r has equation (x − h)² + (y − k)² = r².

Substitution solves a system by replacing one variable; elimination cancels a variable by adding equations.

A system's solution is the point where the two lines intersect on a graph.

Last updated: June 2026

Distance & Midpoint

The distance formula gives the straight-line distance between two points (x₁, y₁) and (x₂, y₂):

d = √((x₂ − x₁)² + (y₂ − y₁)²)

It is simply the Pythagorean theorem applied to the horizontal leg (x₂ − x₁) and vertical leg (y₂ − y₁).

Worked example. Distance between (1, 2) and (4, 6):

d = √((4 − 1)² + (6 − 2)²) = √(3² + 4²) = √(9 + 16) = √25 = 5 .

The midpoint formula gives the point exactly halfway between two points — just average the coordinates:

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Worked example. Midpoint of (−2, 5) and (6, −1):

M = ((−2 + 6)/2, (5 + (−1))/2) = (4/2, 4/2) = (2, 2) .

Formula Use
d = √((x₂−x₁)² + (y₂−y₁)²) length of a segment / distance between points
M = ((x₁+x₂)/2, (y₁+y₂)/2) center of a segment

Common mistake: Forgetting to square both differences, or taking the square root of each term separately — √(9 + 16) = √25 = 5, but √9 + √16 = 3 + 4 = 7, which is wrong.

Circles in the Coordinate Plane

A circle is the set of all points a fixed distance r (the radius) from a center (h, k). Its standard equation is:

(x − h)² + (y − k)² = r²

Read the center by flipping the signs inside the parentheses, and take the square root of the right side for the radius.

Worked example. Identify the circle (x − 3)² + (y + 2)² = 16. The center is (3, −2) (note y + 2 means k = −2), and the radius is √16 = 4 .

Worked example (build the equation). Write the equation of the circle with center (−1, 4) and radius 5. Substitute h = −1, k = 4, r = 5:

(x − (−1))² + (y − 4)² = 5² → (x + 1)² + (y − 4)² = 25 .

Worked example (center and a point). A circle is centered at (0, 0) and passes through (3, 4). The radius is the distance from the center to that point: r = √(3² + 4²) = 5. So the equation is x² + y² = 25.

A circle centered at the origin simplifies to x² + y² = r², because h = k = 0.

Systems of Linear Equations

A system is two (or more) equations solved together; the solution is the (x, y) point that satisfies both — graphically, where the two lines intersect . Parallel lines (no intersection) give no solution ; identical lines give infinitely many .

Substitution — solve one equation for a variable, then substitute into the other.

Worked example. Solve y = 2x − 1 and 3x + y = 9. Substitute the first into the second:

3x + (2x − 1) = 9 → 5x − 1 = 9 → 5x = 10 → x = 2.

Back-substitute: y = 2(2) − 1 = 3. Solution: (2, 3) .

Elimination — add or subtract the equations to cancel one variable, scaling first if needed.

Worked example. Solve 2x + 3y = 12 and 2x − y = 4. Subtract the second from the first:

(2x + 3y) − (2x − y) = 12 − 4 → 4y = 8 → y = 2.

Substitute back: 2x − 2 = 4 → 2x = 6 → x = 3. Solution: (3, 2) .

Method When it's easiest
Substitution one variable is already isolated (e.g., y = …)
Elimination coefficients line up to cancel a variable
Graphing you need the picture / approximate intersection

Common mistake: Solving for only one variable and stopping — a system's answer is an ordered pair (x, y); always find both and check by plugging into both original equations.

Scaling for Elimination & Special Cases

Often the coefficients do not line up to cancel directly, so you multiply one or both equations by a constant first.

Worked example. Solve 3x + 2y = 16 and 5x − 4y = −10. Multiply the first equation by 2 to match the y-terms: 6x + 4y = 32. Now add it to the second:

(6x + 4y) + (5x − 4y) = 32 + (−10) → 11x = 22 → x = 2.

Back-substitute: 3(2) + 2y = 16 → 2y = 10 → y = 5. Solution: (2, 5) .

Recognizing no solution or infinitely many. When you eliminate variables and reach a false statement like 0 = 7, the lines are parallel and there is no solution . When you reach a true statement like 0 = 0, the equations describe the same line and there are infinitely many solutions .

Worked example (no solution). y = 2x + 1 and y = 2x − 3 have equal slopes (2) but different intercepts, so the lines are parallel — no intersection.

Graphical meaning. Each linear equation is a line; the solution is the single crossing point. Two lines either cross once (one solution), never (parallel, none), or overlap completely (same line, infinite). Word problems — mixtures, two-item purchases, distance-rate — translate into systems: define two variables, write one equation per condition, then solve by whichever method is cleaner. Always verify the ordered pair in both original equations before reporting it.