ACT Math: Intermediate Algebra and Coordinate Geometry

TL;DR

This topic combines algebraic skills with geometric concepts, often involving lines, parabolas, circles, and other functions on the coordinate plane. You'll need to know how to manipulate equations, interpret graphs, and find distances, midpoints, and slopes. Focus on understanding the relationships between equations and their visual representations.

1. The Mental Model

Think of intermediate algebra and coordinate geometry as two sides of the same coin: algebra is the language of numbers and symbols, and coordinate geometry is how you visualize that language on a graph. Every equation has a picture, and every picture has an equation.

2. The Core Material

This section covers essential skills you'll need for intermediate algebra and coordinate geometry problems on the ACT.

2.1 Lines and Slopes

A line is defined by its equation, most commonly in slope-intercept form: $y = mx + b$.
* $m$ is the slope, which tells you the steepness and direction of the line. $m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}$.
* $b$ is the y-intercept, the point where the line crosses the y-axis (i.e., when $x=0$, $y=b$).

You should also be familiar with point-slope form: $y - y_1 = m(x - x_1)$, useful when you have a point $(x_1, y_1)$ and the slope $m$.

Parallel lines have the exact same slope.
Perpendicular lines have slopes that are negative reciprocals of each other. If one slope is $m$, the perpendicular slope is $-\frac{1}{m}$.

2.2 Distance and Midpoint Formulas

Given two points $(x_1, y_1)$ and $(x_2, y_2)$:

The distance between them is found using the distance formula, which comes from the Pythagorean theorem:
$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

The midpoint (the exact middle point) between them is:
$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

2.3 Parabolas

Parabolas are graphs of quadratic functions, typically written as $y = ax^2 + bx + c$.
* If $a > 0$, the parabola opens up (like a "U").
* If $a < 0$, the parabola opens down (like an "n").
* The vertex is the highest or lowest point. Its x-coordinate is $x = -\frac{b}{2a}$. Plug this $x$ back into the equation to find the y-coordinate of the vertex.
* The roots (or x-intercepts) are where the parabola crosses the x-axis (where $y=0$). You find these by setting the equation to 0 and solving for $x$ (factoring, quadratic formula, etc.).

2.4 Circles

The standard equation of a circle with center $(h, k)$ and radius $r$ is:
$(x - h)^2 + (y - k)^2 = r^2$

Remember that the $r^2$ in the equation means you need to take the square root of the number on the right side to find the actual radius.

2.5 Interpreting Graphs of Functions

You'll often need to read information directly from a graph. This includes:
* Domain: All possible x-values for which the function is defined (read from left to right).
* Range: All possible y-values the function can output (read from bottom to top).
* Intercepts: Where the graph crosses the x-axis (x-intercepts) or y-axis (y-intercepts).
* Increasing/Decreasing Intervals: Where the graph is going up or down as you move from left to right.

Here's a breakdown of how different coordinate geometry concepts relate:

graph TD
    A["Coordinate Geometry"] --> B["Lines"]
    A --> C["Parabolas"]
    A --> D["Circles"]
    B --> B1["Slope-Intercept Form"]
    B --> B2["Point-Slope Form"]
    B --> B3["Parallel & Perpendicular Slopes"]
    B --> B4["Distance Formula"]
    B --> B5["Midpoint Formula"]
    C --> C1["Vertex (h, k)"]
    C --> C2["Roots (x-intercepts)"]
    C --> C3["Opens Up/Down (from 'a')"]
    D --> D1["Center (h, k)"]
    D --> D2["Radius (r)"]
    D --> D3["Standard Equation"]
    A --> E["Interpreting Graphs"]
    E --> E1["Domain & Range"]
    E --> E2["Intercepts"]
    E --> E3["Increasing/Decreasing"]

3. Worked Example

Problem: A line passes through the points $(2, 5)$ and $(6, 17)$. What is the equation of a line perpendicular to this line that also passes through the point $(4, 3)$?

Solution:

1. Find the slope of the original line:
   $m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{17 - 5}{6 - 2} = \frac{12}{4} = 3$

2. Find the slope of the perpendicular line:
   The slope of the perpendicular line ($m_2$) is the negative reciprocal of $m_1$.
   $m_2 = -\frac{1}{3}$

3. Use the point-slope form for the new line:
   We have the slope $m_2 = -\frac{1}{3}$ and a point $(4, 3)$.
   $y - y_1 = m(x - x_1)$
   $y - 3 = -\frac{1}{3}(x - 4)$

4. Convert to slope-intercept form (optional, but often preferred for answers):
   $y - 3 = -\frac{1}{3}x + \frac{4}{3}$
   $y = -\frac{1}{3}x + \frac{4}{3} + 3$
   $y = -\frac{1}{3}x + \frac{4}{3} + \frac{9}{3}$
   $y = -\frac{1}{3}x + \frac{13}{3}$

So, the equation of the perpendicular line is $y = -\frac{1}{3}x + \frac{13}{3}$.

4. Key Takeaways

• Understand the relationship between algebraic equations and their graphical representations.
• Memorize and know how to apply the slope, distance, and midpoint formulas.
• Be able to identify key features of parabolas (vertex, roots, direction of opening) from their equations.
• Recognize the standard form of a circle's equation and extract its center and radius.
• Practice interpreting graphs to find domain, range, intercepts, and intervals where a function is increasing or decreasing.
• Parallel lines have equal slopes; perpendicular lines have negative reciprocal slopes.
• The quadratic formula is your best friend for finding parabola roots if factoring doesn't work easily.

Common Mistakes to Avoid:
- Mixing up slope and y-intercept when writing linear equations.
- Incorrectly applying the negative reciprocal for perpendicular slopes (forgetting one part or the other).
- Sign errors in distance, midpoint, or circle formulas, especially with negative coordinates.
- Forgetting to square the radius (or take the square root) when working with circle equations.
- Confusing domain (x-values) with range (y-values) when interpreting graphs.

5. Now Try It

Take 15 minutes to solve the following problem:

A circle has its center at $( -1, 2 )$ and passes through the point $( 3, 5 )$.
1. Find the radius of the circle.
2. Write the equation of the circle in standard form.
3. Determine if the point $(0, 6)$ is inside, outside, or on the circle.

What success looks like: You'll accurately use the distance formula to find the radius, then correctly substitute the center and radius into the circle's standard equation. Finally, you'll plug in the test point $(0, 6)$ into your equation to compare the result with $r^2$ and determine its position relative to the circle.