Coordinate Geometry

📍 Coordinate Plane Basics

Part 1 of 7 — Plotting, Quadrants, Distance & Midpoint

The coordinate plane is a two-dimensional surface formed by the intersection of a horizontal number line (the x-axis) and a vertical number line (the y-axis). Every point is described by an ordered pair $(x, y)$.

Key facts:

• Points on the x-axis have $y = 0$.
• Points on the y-axis have $x = 0$.
• The origin is $(0, 0)$.

The Distance Formula

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

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

This comes directly from the Pythagorean theorem applied to the horizontal and vertical legs.

Example 1: Find the distance between $(1, 2)$ and $(4, 6)$.

$d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Example 2: Find the distance between $(-3, 1)$ and $(5, -5)$.

$d = \sqrt{(5-(-3))^2 + (-5-1)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$

ACT Tip: When answer choices are integers, check whether the sum under the radical is a perfect square — on the ACT it often is.

Distance Formula Practice 🎯

The Midpoint Formula

The midpoint of the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is:

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

Simply average the $x$-coordinates and average the $y$-coordinates.

Example 3: Find the midpoint of $(2, 8)$ and $(6, 4)$.

$M = \left(\frac{2+6}{2},\; \frac{8+4}{2}\right) = (4, 6)$

Example 4: The midpoint of $(x, 3)$ and $(7, 11)$ is $(5, 7)$. Find $x$.

$\frac{x + 7}{2} = 5 \implies x + 7 = 10 \implies x = 3$

ACT Tip: The ACT sometimes asks you to find an endpoint given the midpoint and the other endpoint. Use the midpoint formula in reverse: $x_1 = 2M_x - x_2$.

Distance & Midpoint Calculations 🧮

1. Distance between $(3, 0)$ and $(0, 4)$?

2. Midpoint of $(1, 5)$ and $(9, 1)$: what is the x-coordinate?

3. Midpoint of $(-2, 6)$ and $(4, -2)$: what is the y-coordinate?

Quadrant & Formula ID 🔍

ACT-Style Questions 📋

📈 Slope & Linear Equations

Part 2 of 7 — Slope Formula, Slope-Intercept, Point-Slope, Parallel & Perpendicular

The slope of a line through $(x_1, y_1)$ and $(x_2, y_2)$ is:

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

Linear equation forms:

• Slope-intercept: $y = mx + b$ (slope $m$, y-intercept $b$)
• Point-slope: $y - y_1 = m(x - x_1)$
• Standard form: $Ax + By = C$

Worked Examples

Example 1 — Slope: Find the slope through $(2, 3)$ and $(6, 11)$.

$m = \frac{11 - 3}{6 - 2} = \frac{8}{4} = 2$

Example 2 — Slope-intercept: A line has slope $3$ and y-intercept $-5$. Write its equation.

$y = 3x - 5$

Example 3 — Point-slope: Write the equation of the line through $(4, 1)$ with slope $-2$.

$y - 1 = -2(x - 4) \implies y = -2x + 9$

Parallel & Perpendicular:

• Parallel lines have the same slope: $m_1 = m_2$.
• Perpendicular lines have negative reciprocal slopes: $m_1 \cdot m_2 = -1$.

Example 4: A line has slope $\frac{3}{4}$. A perpendicular line has slope $-\frac{4}{3}$.

ACT Tip: Check a perpendicular slope by multiplying: the product must be $-1$. A choice with the same slope is parallel, and a choice with only the sign flipped (or only the fraction flipped) is a trap — eliminate them quickly.

Slope & Equations 🎯

Slope Calculations 🧮

1. Slope through $(0, 4)$ and $(2, 10)$?

2. y-intercept of $y = -3x + 7$? (just the number)

3. If a line has slope $5$, its perpendicular has slope $-1/k$. What is $k$?

Line Relationships 🔍

Perpendicular Lines — Full Example

Problem: Find the equation of the line perpendicular to $y = \frac{2}{3}x + 4$ that passes through $(6, 1)$.

Step 1: The given slope is $\frac{2}{3}$. The perpendicular slope is $m = -\frac{3}{2}$.

Step 2: Use point-slope form with $(6, 1)$:

$y - 1 = -\frac{3}{2}(x - 6)$

$y - 1 = -\frac{3}{2}x + 9$

$y = -\frac{3}{2}x + 10$

Perpendicular bisector: the perpendicular bisector of a segment is the line that passes through the segment's midpoint and is perpendicular to it. Find the midpoint, then use the negative reciprocal of the segment's slope.

Example: For the segment from $(1, 2)$ to $(5, 6)$: midpoint $= (3, 4)$, segment slope $= \frac{6-2}{5-1} = 1$, so the bisector has slope $-1$: $y - 4 = -1(x - 3) \implies y = -x + 7$.

ACT Tip: Convert to slope-intercept form ($y = mx + b$) to match answer choices quickly.

ACT-Style Questions 📋

📊 Graphing Lines & Inequalities

Part 3 of 7 — Intercepts, Graphing Methods, Shading Regions

There are three standard ways to graph a line:

Finding intercepts:

• x-intercept: Set $y = 0$ and solve for $x$.
• y-intercept: Set $x = 0$ and solve for $y$.

Example 1: $3x + 2y = 12$

• x-intercept: $3x = 12 \implies x = 4$ → point $(4, 0)$
• y-intercept: $2y = 12 \implies y = 6$ → point $(0, 6)$

Graphing with Slope-Intercept Form

Given $y = mx + b$:

1. Plot the y-intercept $(0, b)$.
2. From that point, use the slope $m = \frac{\text{rise}}{\text{run}}$ to find the next point.
3. Draw the line through both points.

Example 2: Graph $y = -\frac{2}{3}x + 4$.

• Start at $(0, 4)$.
• Slope $= -\frac{2}{3}$: go down 2, right 3 → $(3, 2)$.
• Draw a line through $(0, 4)$ and $(3, 2)$.

Inequalities change two things:

• $<$ or $>$: dashed line (boundary NOT included).
• $\le$ or $\ge$: solid line (boundary included).
• Shade above the line for $y > mx + b$ or $y \ge mx + b$.
• Shade below the line for $y < mx + b$ or $y \le mx + b$.

ACT Tip: To check which side to shade, test the point $(0, 0)$. If it satisfies the inequality, shade the side containing the origin; if not, shade the other side. (If the line passes through the origin, test a different point such as $(1, 0)$.)

Intercepts & Graphing 🎯

Finding Intercepts 🧮

For each equation, find the requested intercept value.

1. y-intercept of $4x + y = 9$? (just the y-value)

2. x-intercept of $2x - 6y = 18$? (just the x-value)

3. y-intercept of $y = 5x - 15$? (just the y-value)

Inequality Graphing 🔍

Systems of Inequalities

When two inequalities are graphed together, the solution region is where the shading overlaps.

Example 3: Graph the system:

$y \ge x - 1$ $y < -2x + 5$

• First inequality: solid line through $(0, -1)$ with slope $1$; shade above.
• Second inequality: dashed line through $(0, 5)$ with slope $-2$; shade below.
• The solution is the region that satisfies both — the overlap area.

ACT Tip: On the ACT, they often ask which point is in the solution region. Plug each answer choice into both inequalities — the correct answer satisfies both.

ACT-Style Questions 📋

⭕ Circles on the Coordinate Plane

Part 4 of 7 — Standard Form, Center & Radius, Completing the Square

The standard form equation of a circle is:

$(x - h)^2 + (y - k)^2 = r^2$

• Center: $(h, k)$
• Radius: $r$
• Diameter: $2r$ (so the radius is half the diameter)

If a problem gives the endpoints of a diameter, the center is their midpoint and the radius is half the distance between them.

Key insight: Watch the signs! $(y + 1)$ means $k = -1$, and $(x + 2)$ means $h = -2$.

Completing the Square for Circles

The ACT may give a circle in general form:

$x^2 + y^2 + Dx + Ey + F = 0$

Convert it by completing the square for both variables.

Example 1: Rewrite $x^2 + y^2 - 8x + 6y - 11 = 0$.

Step 1: Group terms: $(x^2 - 8x) + (y^2 + 6y) = 11$

Step 2: Complete each square:

• $x$: $\left(\frac{-8}{2}\right)^2 = 16$
• $y$: $\left(\frac{6}{2}\right)^2 = 9$

Step 3: Add to both sides:

$(x^2 - 8x + 16) + (y^2 + 6y + 9) = 11 + 16 + 9$

$(x - 4)^2 + (y + 3)^2 = 36$

Center $= (4, -3)$, radius $= 6$. ✓

ACT Tip: On the ACT, you usually just need the center or radius — focus on completing the square correctly rather than graphing.

Circle Equations 🎯

Circle Calculations 🧮

Given $x^2 + y^2 - 4x + 10y + 20 = 0$, complete the square.

1. What is the x-coordinate of the center?

2. What is the y-coordinate of the center?

3. What is the radius?

Circle Properties 🔍

Tangent Lines & Point-on-Circle Problems

Does a point lie on a circle? Substitute it into the equation and check.

Example 2: Does $(3, 4)$ lie on $x^2 + y^2 = 25$?

$3^2 + 4^2 = 9 + 16 = 25 \; ✓$

Yes, $(3, 4)$ is on the circle.

Example 3: Does $(1, 5)$ lie on $(x - 2)^2 + (y + 1)^2 = 36$?

$(1 - 2)^2 + (5 + 1)^2 = 1 + 36 = 37 \neq 36$

No — $37 > 36$ means $(1,5)$ is outside the circle.

ACT Tip: This substitution test is fast and appears frequently on the ACT. No need for completing the square if the equation is already in standard form.

ACT-Style Questions 📋

🔵 Conic Sections Overview

Part 5 of 7 — Parabola Vertex Form, Ellipses & Hyperbolas for the ACT

A conic section is a curve obtained by slicing a cone with a plane. The four types are:

On the ACT, parabolas appear most often. Ellipses and hyperbolas are rare but worth recognizing.

Parabolas in Vertex Form

The vertex form of a parabola is:

$y = a(x - h)^2 + k$

• Vertex: $(h, k)$ — the highest or lowest point.
• If $a > 0$: opens upward (vertex is a minimum).
• If $a < 0$: opens downward (vertex is a maximum).
• $|a|$ controls the width: larger $|a|$ = narrower parabola.

Example 1: $y = 2(x - 3)^2 + 1$

• Vertex: $(3, 1)$
• Opens up (since $a = 2 > 0$)
• Narrower than $y = x^2$ (since $|a| = 2 > 1$)

Example 2: $y = -(x + 4)^2 + 9$

• Vertex: $(-4, 9)$
• Opens down (since $a = -1 < 0$)
• Maximum value is $y = 9$

Axis of symmetry: $x = h$ (vertical line through the vertex).

ACT Tip: The vertex tells you the max/min value immediately — no calculus needed!

Parabola Properties 🎯

Ellipses & Hyperbolas (ACT Basics)

Ellipse: $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$

• Center: $(h, k)$
• The larger denominator determines the major axis direction.
• If $a^2 > b^2$: horizontal major axis (wider).
• If $b^2 > a^2$: vertical major axis (taller).

Example 3: $\frac{x^2}{25} + \frac{y^2}{9} = 1$

• Center: $(0, 0)$, $a = 5$, $b = 3$.
• Horizontal major axis, stretches $5$ units left/right and $3$ units up/down.

Hyperbola: $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$

• Note the minus sign — this distinguishes it from an ellipse.
• Opens left and right when the $x$-term is positive.
• Opens up and down when the $y$-term is positive.

ACT Tip: On the ACT, you mainly need to identify the conic type and find the center/vertex. Deep analysis is rare.

Conic Section Identification 🧮

1. Vertex x-coordinate of $y = -(x - 6)^2 + 2$?

2. Vertex y-coordinate of $y = -(x - 6)^2 + 2$?

3. For $\frac{x^2}{16} + \frac{y^2}{4} = 1$, what is the value of $a$ (the larger semi-axis)?

Conic Type Identification 🔍

ACT-Style Questions 📋

🔄 Transformations on the Coordinate Plane

Part 6 of 7 — Translations, Reflections, Rotations & Dilations

A transformation changes a figure's position, size, or orientation. The four main types:

Rigid motions (translation, reflection, rotation) preserve distances and angles.

Translations (Slides)

A translation shifts every point by the same amount.

$(x, y) \to (x + a,\; y + b)$

• $a > 0$: shift right. $a < 0$: shift left.
• $b > 0$: shift up. $b < 0$: shift down.

Example 1: Translate $(3, -2)$ by $\langle -4, 5 \rangle$.

$(3 + (-4),\; -2 + 5) = (-1, 3)$

Reflections (Flips)

Example 2: Reflect $(4, -7)$ over the x-axis → $(4, 7)$.

Example 3: Reflect $(-3, 5)$ over the y-axis → $(3, 5)$.

ACT Tip: Reflection over the x-axis flips the $y$-sign. Reflection over the y-axis flips the $x$-sign. Just remember which coordinate changes.

Translations & Reflections 🎯

Rotations about the Origin

Example 4: Rotate $(3, 5)$ by $90°$ counterclockwise → $(-5, 3)$.

Example 5: Rotate $(-2, 4)$ by $180°$ → $(2, -4)$.

Dilations (Resizing)

A dilation with center at the origin and scale factor $k$:

$(x, y) \to (kx, ky)$

• $k > 1$: enlargement.
• $0 < k < 1$: reduction.
• $k = 1$: no change.

Example 6: Dilate $(4, -6)$ by scale factor $\frac{1}{2}$ → $(2, -3)$.

ACT Tip: After a dilation by factor $k$, distances are multiplied by $|k|$ and areas are multiplied by $k^2$.

Transformation Calculations 🧮

1. Rotate $(4, 2)$ by $90°$ counterclockwise. What is the new x-coordinate?

2. Reflect $(7, -3)$ over the line $y = x$. What is the new x-coordinate?

3. Dilate $(6, 9)$ by scale factor $\frac{1}{3}$. What is the new y-coordinate?

Transformation Types 🔍

ACT-Style Questions 📋

🏆 Review & Mixed Practice

Part 7 of 7 — Formula Cheat Sheet & Mixed ACT Coordinate Geometry Problems

Here is your complete cheat sheet of coordinate geometry formulas for the ACT:

Strategy for ACT Coordinate Geometry:

1. Identify what formula you need.
2. Label known values clearly.
3. Plug in and simplify.
4. Watch for sign errors — they are one of the most common mistakes.

Quick Review — Key Concepts

Quadrants: Signs of $(x, y)$ — I: $(+,+)$, II: $(-,+)$, III: $(-,-)$, IV: $(+,-)$.

Slope ideas:

• Horizontal line: $m = 0$, equation $y = c$.
• Vertical line: $m$ undefined, equation $x = c$.
• Parallel $\implies$ same slope.
• Perpendicular $\implies$ negative reciprocal slopes.

Circles: Complete the square to go from general to standard form. Center and radius come directly from $(x-h)^2 + (y-k)^2 = r^2$.

Transformations summary:

ACT Tip: The Enhanced ACT Math section gives you 50 minutes for 45 questions (4 answer choices each), so you average just over a minute per question. Don't derive formulas — memorize them!

Mixed Review — Set 1 🎯

Mixed Calculations 🧮

1. Slope of the line through $(2, 5)$ and $(8, 17)$?

2. The midpoint of $(0, 0)$ and $(10, 6)$: what is the x-coordinate?

3. A circle has equation $(x-1)^2 + (y+3)^2 = 64$. What is the radius?

Formula Matching 🔍

Mixed ACT-Style Practice

Try these without a calculator — ACT coordinate geometry usually involves clean numbers.

ACT Tip: On test day, write down the formulas you've memorized before starting. This saves time and reduces errors under pressure.

ACT-Style Questions — Final Set 📋