2.5 Coordinate Geometry, Lines, and Functions

Coordinate geometry connects algebraic equations with geometric visualizations on the two-dimensional Cartesian coordinate plane. Understanding lines, slopes, distances, function mappings, domain/range constraints, and function compositions is essential for solving spatial and analytical quantitative problems on the AdUCET.

1. The Cartesian Plane and Fundamental Formulas

The Cartesian coordinate plane is formed by two perpendicular axes intersecting at the origin $(0,0)$: the horizontal $x$-axis and the vertical $y$-axis. Any point in the plane is represented by an ordered pair $(x, y)$.

The Distance Formula

Derived directly from the Pythagorean Theorem ($a^2 + b^2 = c^2$), the distance $d$ between two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ is:

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

The Midpoint Formula

The midpoint $M$ of the line segment connecting $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ represents the average of their respective $x$- and $y$-coordinates:

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

Worked Step-by-Step Example

Find the distance $d$ and midpoint $M$ between points $A(-2, 5)$ and $B(4, -3)$.

1. Distance Calculation: $$d = \sqrt{(4 - (-2))^2 + (-3 - 5)^2} = \sqrt{(6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$
2. Midpoint Calculation: $$M = \left( \frac{-2 + 4}{2}, \frac{5 + (-3)}{2} \right) = \left( \frac{2}{2}, \frac{2}{2} \right) = (1, 1)$$

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2. Slope of a Line

The slope $m$ measures the steepness and direction of a line, defined as the ratio of vertical change ("rise") to horizontal change ("run") between any two distinct points $(x_1, y_1)$ and $(x_2, y_2)$:

$$m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} \quad (x_1 \neq x_2)$$

Four Slope Orientations

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3. Standard Forms of Linear Equations

Linear equations can be expressed in three main algebraic formats:

Finding Line Intercepts

• $x$-intercept: Set $y = 0$ and solve for $x$. The line crosses the horizontal axis at $(x_{\text{int}}, 0)$.
• $y$-intercept: Set $x = 0$ and solve for $y$. The line crosses the vertical axis at $(0, y_{\text{int}})$.

Worked Example: Intercepts from Standard Form

Given $3x - 4y = 12$:

• To find $x$-intercept: $3x - 4(0) = 12 \implies 3x = 12 \implies x = 4$. Point is $(4,0)$.
• To find $y$-intercept: $3(0) - 4y = 12 \implies -4y = 12 \implies y = -3$. Point is $(0,-3)$.

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4. Parallel and Perpendicular Lines

The geometric relationship between two lines on the Cartesian plane is dictated entirely by their slopes $m_1$ and $m_2$.

Parallel Lines ($m_1 = m_2$)

Parallel lines run infinitely in the same direction without intersecting. They have identical slopes and distinct $y$-intercepts:

$$m_1 = m_2 \quad (b_1 \neq b_2)$$

Perpendicular Lines ($m_1 \cdot m_2 = -1$)

Perpendicular lines intersect at right angles ($90^\circ$). Their slopes are negative reciprocals of each other:

$$m_2 = -\frac{1}{m_1} \implies m_1 \cdot m_2 = -1$$

Worked Example: Constructing a Perpendicular Line

Find the equation of the line passing through point $P(3, -2)$ that is perpendicular to the line $y = \frac{1}{2}x + 4$.

1. Identify Given Slope: The given line has slope $m_1 = \frac{1}{2}$.
2. Determine Perpendicular Slope: $m_2 = -\frac{1}{1/2} = -2$.
3. Apply Point-Slope Form: $$y - (-2) = -2(x - 3) \implies y + 2 = -2x + 6 \implies y = -2x + 4$$

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5. Functions, Domain, and Range

A function $f$ is a binary relation assigning to each element $x$ in a set called the domain exactly one element $y = f(x)$ in a set called the range.

The Vertical Line Test

A graph in the coordinate plane represents a function if and only if no vertical line intersects the curve at more than one point.

Domain Restrictions

When determining the implicit domain of a real-valued algebraic function, apply two main rules:

1. Division by Zero: Denominators cannot equal zero.
   Example: For $f(x) = \frac{5}{x - 3}$, set $x - 3 \neq 0 \implies \text{Domain: } x \neq 3$ or $(-\infty, 3) \cup (3, \infty)$.
2. Even Radicands: Expressions inside even roots (e.g., square roots) must be non-negative.
   Example: For $g(x) = \sqrt{2x - 8}$, set $2x - 8 \ge 0 \implies 2x \ge 8 \implies \text{Domain: } x \ge 4$ or $[4, \infty)$.

Determining Function Range

The range consists of all possible output values $y$ produced by $f(x)$ across its domain:

• Linear Functions ($f(x) = mx + b, m \neq 0$): Range is all real numbers $(-\infty, \infty)$.
• Quadratic Functions ($f(x) = ax^2 + bx + c$): Range depends on the vertex $y$-coordinate $k = f\left(-\frac{b}{2a}\right)$. If $a > 0$, range is $[k, \infty)$; if $a < 0$, range is $(-\infty, k]$.
• Absolute Value Functions ($f(x) = |x| + k$): Range is $[k, \infty)$.

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6. Function Evaluation and Composition

Evaluating Functions

Evaluating $f(a)$ means replacing every instance of the variable $x$ in $f(x)$ with the input value or expression $a$.

Composite Functions

Given two functions $f(x)$ and $g(x)$, the composite function $(f \circ g)(x)$, read as "$f$ of $g$ of $x$", is defined by:

$$(f \circ g)(x) = f(g(x))$$

The output of the inner function $g(x)$ becomes the direct input for the outer function $f(x)$.

Worked Step-by-Step Example

Let $f(x) = x^2 - 3x + 2$ and $g(x) = 2x + 1$.

1. Evaluate $g(3)$: $$g(3) = 2(3) + 1 = 7$$
2. Evaluate $(f \circ g)(3) = f(g(3)) = f(7)$: $$f(7) = (7)^2 - 3(7) + 2 = 49 - 21 + 2 = 30$$
3. Find Algebraic Expression for $(f \circ g)(x)$: $$(f \circ g)(x) = f(2x + 1) = (2x + 1)^2 - 3(2x + 1) + 2$$ $$= (4x^2 + 4x + 1) - (6x + 3) + 2 = 4x^2 - 2x$$