Coordinate Geometry

Learn slope, distance, midpoint, and line equations for ACT coordinate geometry with formulas, examples, and practice problems.

Coordinate geometry bridges algebra and geometry by placing shapes, lines, and curves on the xyxy-plane. On the ACT Math section, coordinate geometry accounts for roughly 8–10 questions out of 60, making it one of the most consistently tested domains on the exam. These questions range from straightforward slope calculations to multi-step problems involving the distance formula, midpoint, parallel and perpendicular lines, and circles.

The ACT Math section gives you 60 questions in 60 minutes — approximately one minute per question — with a calculator allowed throughout. Unlike the SAT, there is no separate "no calculator" section. There is also no penalty for wrong answers, so always guess if you are unsure. Coordinate geometry questions tend to appear in the middle portion of the test (roughly questions 20–45), placing them in the medium-difficulty range. If you know the formulas and can apply them quickly, these are very achievable points.

One critical difference between the ACT and SAT: the ACT does not provide a formula reference sheet. You must have the slope formula, distance formula, midpoint formula, and circle equation memorized before test day. There is simply no way around this requirement.

This guide covers every coordinate geometry concept the ACT tests: slope and equations of lines, the distance and midpoint formulas, parallel and perpendicular relationships, graphing, and circles in the coordinate plane. By the end, you will be able to tackle any coordinate geometry question with confidence and speed.

Core Concepts

The Coordinate Plane

The coordinate plane consists of two perpendicular number lines: the horizontal xx-axis and the vertical yy-axis. They intersect at the origin (0,0)(0, 0). Every point is described by an ordered pair (x,y)(x, y), where xx is the horizontal distance from the origin and yy is the vertical distance.

The four quadrants are:

  • Quadrant I: x>0x > 0, y>0y > 0 (upper right)
  • Quadrant II: x<0x < 0, y>0y > 0 (upper left)
  • Quadrant III: x<0x < 0, y<0y < 0 (lower left)
  • Quadrant IV: x>0x > 0, y<0y < 0 (lower right)

The ACT may ask which quadrant a point lies in, or which quadrants a line passes through. Knowing the sign conventions is essential.

Slope

The slope of a line measures its steepness and direction. Given two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2):

m=y2y1x2x1=riserunm = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}

Key slope facts every ACT student must know:

  • Positive slope: line rises from left to right (goes uphill)
  • Negative slope: line falls from left to right (goes downhill)
  • Zero slope: horizontal line (y=cy = c) — the rise is zero
  • Undefined slope: vertical line (x=cx = c) — the run is zero (division by zero)

The slope tells you how much yy changes for every 1-unit change in xx. A slope of 34\frac{3}{4} means "up 3, right 4."

Equations of Lines

Slope-intercept form: y=mx+by = mx + b, where mm is the slope and bb is the yy-intercept (the point where the line crosses the yy-axis). This is the most useful form for the ACT because it immediately reveals both the slope and the intercept.

Point-slope form: yy1=m(xx1)y - y_1 = m(x - x_1) — useful when you know a specific point (x1,y1)(x_1, y_1) and the slope mm. You can always convert this to slope-intercept form by distributing and simplifying.

Standard form: Ax+By=CAx + By = C, where AA, BB, and CC are integers. The ACT often presents equations this way. To find the slope from standard form, rearrange to slope-intercept form or use the shortcut: m=ABm = -\frac{A}{B}.

Finding the equation of a line: The most common ACT question type asks you to find the equation of a line given two points, or a point and a slope.

  1. Find the slope using m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}
  2. Plug the slope and one point into point-slope form
  3. Simplify to slope-intercept form if needed

The Distance Formula

The distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is:

d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

This formula is simply the Pythagorean theorem applied to the coordinate plane. The horizontal distance x2x1|x_2 - x_1| and vertical distance y2y1|y_2 - y_1| form the legs of a right triangle, and the straight-line distance is the hypotenuse. If you ever forget the distance formula, draw the right triangle and use a2+b2=c2a^2 + b^2 = c^2.

Pro tip: Before computing the full square root, check if the numbers form a Pythagorean triple. If the legs are 6 and 8, the hypotenuse is 10 (the 33-44-55 triple scaled by 2). Recognizing triples saves significant time.

The Midpoint Formula

The midpoint of the segment connecting (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is:

M=(x1+x22,y1+y22)M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

Think of it as averaging the xx-coordinates and averaging the yy-coordinates. The midpoint is exactly halfway between the two points in both directions.

A useful extension: if you know one endpoint and the midpoint, you can find the other endpoint. If (x1,y1)(x_1, y_1) and midpoint (mx,my)(m_x, m_y) are known, then x2=2mxx1x_2 = 2m_x - x_1 and y2=2myy1y_2 = 2m_y - y_1.

Parallel and Perpendicular Lines

These relationships are among the most frequently tested concepts in ACT coordinate geometry:

  • Parallel lines have the same slope: m1=m2m_1 = m_2. They never intersect.
  • Perpendicular lines have slopes that are negative reciprocals: m1m2=1m_1 \cdot m_2 = -1, or equivalently m2=1m1m_2 = -\frac{1}{m_1}.

For example, if a line has slope 23\frac{2}{3}:

  • A parallel line also has slope 23\frac{2}{3}
  • A perpendicular line has slope 32-\frac{3}{2} (flip the fraction and negate)

Special cases: A horizontal line (m=0m = 0) is perpendicular to a vertical line (undefined slope).

Intercepts

The xx-intercept is the point where the line crosses the xx-axis. At this point, y=0y = 0. To find it, set y=0y = 0 in the equation and solve for xx.

The yy-intercept is the point where the line crosses the yy-axis. At this point, x=0x = 0. To find it, set x=0x = 0 (or read the value of bb directly from y=mx+by = mx + b).

Graphing Linear Equations

To graph y=mx+by = mx + b:

  1. Plot the yy-intercept (0,b)(0, b)
  2. Use the slope m=riserunm = \frac{\text{rise}}{\text{run}} to find a second point (from the yy-intercept, go up "rise" units and right "run" units)
  3. Draw the line through both points

For horizontal lines y=cy = c: draw a horizontal line at height cc.

For vertical lines x=cx = c: draw a vertical line at position cc.

Circles in the Coordinate Plane

The standard equation of a circle with center (h,k)(h, k) and radius rr is:

(xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2

Watch the signs: In (x3)2+(y+1)2=25(x - 3)^2 + (y + 1)^2 = 25, the center is (3,1)(3, -1) and the radius is 25=5\sqrt{25} = 5. The +1+1 inside the parentheses with yy means k=1k = -1 (the sign flips).

To determine whether a point lies inside, on, or outside a circle, compute the distance from the point to the center and compare it to the radius:

  • If distance <r< r: inside
  • If distance =r= r: on the circle
  • If distance >r> r: outside

Strategy Tips

Tip 1: Sketch It Out

Many coordinate geometry problems become dramatically easier with a quick sketch. Even a rough diagram on your test booklet helps you visualize relationships, estimate answers, and eliminate wrong choices. Spend 10 seconds sketching — it can save you minutes of confusion.

Tip 2: Memorize the Formulas

The ACT does not provide a formula sheet. You must have slope, distance, midpoint, and the circle equation memorized cold before test day. Consider writing them on your scratch paper in the first 30 seconds of the section.

Tip 3: Use the Answer Choices to Estimate

If the question asks for the distance between two points and the choices are 5, 10, 13, 15, and 17, you can sometimes estimate by counting grid squares on your sketch rather than doing the full calculation. This is especially helpful when you are short on time.

Tip 4: Convert to Slope-Intercept Form Immediately

When an equation is given in standard form (Ax+By=CAx + By = C), your first move should be converting to y=mx+by = mx + b. This takes a few seconds and makes slope, intercept, and comparison problems trivial.

Tip 5: Remember the Special Relationships

Parallel = same slope. Perpendicular = negative reciprocal. These two facts alone can answer many ACT questions in under 30 seconds. Drill them until they are automatic.

Worked Example: Example 1

Problem

What is the slope of the line passing through (2,5)(-2, 5) and (4,1)(4, -1)?

Solution

m=154(2)=66=1m = \frac{-1 - 5}{4 - (-2)} = \frac{-6}{6} = -1

The slope is 1-1.

Worked Example: Example 2

Problem

What is the distance between the points (1,3)(1, 3) and (7,11)(7, 11)?

Solution

d=(71)2+(113)2=36+64=100=10d = \sqrt{(7-1)^2 + (11-3)^2} = \sqrt{36 + 64} = \sqrt{100} = 10

Notice: the legs are 6 and 8, forming the classic 66-88-1010 Pythagorean triple.

Worked Example: Example 3

Problem

Line \ell passes through (2,5)(2, 5) and is perpendicular to the line y=13x+4y = \frac{1}{3}x + 4. What is the equation of line \ell?

Solution

The given line has slope 13\frac{1}{3}. A perpendicular line has slope m=3m = -3 (negative reciprocal).

Using point-slope form with the point (2,5)(2, 5): y5=3(x2)y - 5 = -3(x - 2) y5=3x+6y - 5 = -3x + 6 y=3x+11y = -3x + 11

Worked Example: Example 4

Problem

What is the midpoint of the segment with endpoints (3,8)(-3, 8) and (7,2)(7, -2)?

Solution

M=(3+72,8+(2)2)=(42,62)=(2,3)M = \left(\frac{-3 + 7}{2}, \frac{8 + (-2)}{2}\right) = \left(\frac{4}{2}, \frac{6}{2}\right) = (2, 3)

Worked Example: Example 5

Problem

A circle has the equation (x3)2+(y+1)2=25(x - 3)^2 + (y + 1)^2 = 25. Does the point (6,3)(6, 3) lie inside, on, or outside the circle?

Solution

The center is (3,1)(3, -1) and the radius is 25=5\sqrt{25} = 5.

Calculate the distance from the center to (6,3)(6, 3): d=(63)2+(3(1))2=9+16=25=5d = \sqrt{(6-3)^2 + (3-(-1))^2} = \sqrt{9 + 16} = \sqrt{25} = 5

Since d=r=5d = r = 5, the point lies exactly on the circle.

Worked Example: Example 6

Problem

The endpoints of a diameter of a circle are (2,1)(2, -1) and (8,7)(8, 7). What is the equation of the circle?

Solution

The center is the midpoint of the diameter: C=(2+82,1+72)=(5,3)C = \left(\frac{2+8}{2}, \frac{-1+7}{2}\right) = (5, 3)

The radius is half the length of the diameter. First find the diameter length: d=(82)2+(7(1))2=36+64=100=10d = \sqrt{(8-2)^2 + (7-(-1))^2} = \sqrt{36 + 64} = \sqrt{100} = 10

So r=5r = 5 and the equation is (x5)2+(y3)2=25(x-5)^2 + (y-3)^2 = 25.

Practice Problems

  1. Problem 1

    What is the slope of a line parallel to 4x+2y=104x + 2y = 10?

    A) 2-2 \quad B) 22 \quad C) 12-\frac{1}{2} \quad D) 12\frac{1}{2} \quad E) 55

    Answer: A) 2-2. Rearrange: 2y=4x+102y = -4x + 10, so y=2x+5y = -2x + 5. The slope is 2-2; parallel lines have the same slope.

    Problem 2

    What is the midpoint of the segment from (0,0)(0, 0) to (8,6)(8, 6)?

    A) (8,6)(8, 6) \quad B) (4,3)(4, 3) \quad C) (16,12)(16, 12) \quad D) (3,4)(3, 4) \quad E) (2,3)(2, 3)

    Answer: B) (4,3)(4, 3). Midpoint =(0+82,0+62)=(4,3)= (\frac{0+8}{2}, \frac{0+6}{2}) = (4, 3).

    Problem 3

    What is the distance between (1,2)(-1, 2) and (5,10)(5, 10)?

    A) 88 \quad B) 1010 \quad C) 1414 \quad D) 52\sqrt{52} \quad E) 66

    Answer: B) 1010. d=(5(1))2+(102)2=36+64=100=10d = \sqrt{(5-(-1))^2 + (10-2)^2} = \sqrt{36+64} = \sqrt{100} = 10.

    Problem 4

    A line passes through (0,4)(0, -4) and is perpendicular to a line with slope 25\frac{2}{5}. What is the equation of this line?

    A) y=52x4y = -\frac{5}{2}x - 4 \quad B) y=25x4y = \frac{2}{5}x - 4 \quad C) y=52x4y = \frac{5}{2}x - 4 \quad D) y=25x4y = -\frac{2}{5}x - 4 \quad E) y=52x+4y = -\frac{5}{2}x + 4

    Answer: A) y=52x4y = -\frac{5}{2}x - 4. The perpendicular slope is 52-\frac{5}{2}, and the line passes through (0,4)(0, -4), so b=4b = -4.

    Problem 5

    A circle centered at (2,3)(-2, 3) passes through the point (1,7)(1, 7). What is the radius of the circle?

    A) 33 \quad B) 44 \quad C) 55 \quad D) 77 \quad E) 2525

    Answer: C) 55. r=(1(2))2+(73)2=9+16=25=5r = \sqrt{(1-(-2))^2 + (7-3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.

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Common Mistakes

  • Swapping xx and yy in the slope formula. Remember: slope is y2y1x2x1\frac{y_2 - y_1}{x_2 - x_1} (rise over run), not x2x1y2y1\frac{x_2 - x_1}{y_2 - y_1} (run over rise). This is one of the most common errors.
  • Forgetting the negative reciprocal for perpendicular lines. Students often just take the reciprocal without negating, or negate without taking the reciprocal. You must do both: flip and negate.
  • Arithmetic errors under the square root. In the distance formula, be careful with signs: (35)2=(8)2=64(-3 - 5)^2 = (-8)^2 = 64, not 64-64. Squaring always gives a positive result.
  • Mixing up center signs in circle equations. In (x3)2+(y+1)2=25(x - 3)^2 + (y + 1)^2 = 25, the center is (3,1)(3, -1), not (3,1)(3, 1). The sign inside the parentheses is always the opposite of the coordinate.
  • Confusing xx-intercept and yy-intercept. The xx-intercept is found by setting y=0y = 0; the yy-intercept is found by setting x=0x = 0. Getting these backwards leads to wrong answers.
  • Using diameter instead of radius (or vice versa) in circle equations. The circle equation uses r2r^2, where rr is the radius. If a problem gives you the diameter, divide by 2 first.

Frequently Asked Questions

Is the distance formula just the Pythagorean theorem?

Yes, exactly. The distance formula is derived from the Pythagorean theorem applied to a right triangle formed by two points and their horizontal/vertical distances. If you forget the distance formula, draw a right triangle on the coordinate plane and use a2+b2=c2a^2 + b^2 = c^2.

How do I know when to use point-slope form vs. slope-intercept form?

Use point-slope form when you are given (or can find) a specific point and a slope. Use slope-intercept form when you need to compare slopes and intercepts or when the problem asks for the equation in y=mx+by = mx + b form. On the ACT, converting everything to slope-intercept form is usually the safest bet.

How many coordinate geometry questions are on the ACT?

Typically 8–10 out of 60 questions. The ACT officially categorizes about 15% of questions as coordinate geometry, making it one of the five main content categories.

Do I need to know about conic sections beyond circles?

The ACT occasionally tests parabolas in vertex form (y=a(xh)2+ky = a(x-h)^2 + k), but ellipses and hyperbolas are extremely rare (perhaps once in several tests). Focus your study time on circles and lines for the biggest payoff.

Can I use my graphing calculator for coordinate geometry?

Absolutely. Graphing calculators can plot lines, find intersections, calculate distances, and even graph circles. Use your calculator to verify your algebraic work, especially on problems where you are not 100% confident.

Key Takeaways

  • Memorize the three key formulas. Slope (y2y1x2x1\frac{y_2-y_1}{x_2-x_1}), distance ((x2x1)2+(y2y1)2\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}), and midpoint (x1+x22,y1+y22\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) appear on nearly every ACT, and no formula sheet is provided.

  • Parallel lines share slopes; perpendicular lines have negative reciprocal slopes. These two relationships are tested repeatedly and are among the easiest concepts to apply once memorized.

  • Convert to slope-intercept form. When given standard form (Ax+By=CAx + By = C), rearranging to y=mx+by = mx + b makes slope and intercept immediately visible.

  • Sketch the coordinate plane. A quick diagram prevents sign errors, clarifies relationships, and helps eliminate unreasonable answers.

  • The distance formula is the Pythagorean theorem. If you forget the formula, draw the right triangle and compute c=a2+b2c = \sqrt{a^2 + b^2}. Know common triples (33-44-55, 55-1212-1313).

  • Watch signs in circle equations. The center of (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2 is (h,k)(h, k) — the signs inside the parentheses are opposite to the coordinates.

  • Budget your time wisely. Coordinate geometry questions are medium difficulty on average. Aim for about one minute each, and do not let any single problem consume more than 90 seconds.

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