Lines and Circles: Equations, Graphs, and Applications

Equations of Lines

Forms of Linear Equations

Understanding the different forms of linear equations is fundamental in Precalculus. The most common forms are:

• Slope-Intercept Form: $y = mx + b$, where m is the slope and b is the y-intercept.

• Point-Slope Form: $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and m is the slope.

• Standard Form: $Ax + By = C$, where A, B, and C are constants.

Key Point: The slope m measures the steepness of the line and is calculated as $m = \frac{y_2 - y_1}{x_2 - x_1}$ for two points $(x_1, y_1)$ and $(x_2, y_2)$.

Example: Find the equation of the line passing through $(2, 3)$ and $(4, 7)$.

• Calculate the slope: $m = \frac{7 - 3}{4 - 2} = 2$

• Use point-slope form: $y - 3 = 2(x - 2)$

• Simplify to slope-intercept form: $y = 2x - 1$

Parallel and Perpendicular Lines

• Parallel lines have the same slope: $m_1 = m_2$.

• Perpendicular lines have slopes that are negative reciprocals: $m_1 \cdot m_2 = -1$.

Example: Find the equation of the line perpendicular to $y = 3x + 2$ passing through $(1, -4)$.

• Slope of given line: $m = 3$

• Slope of perpendicular line: $m = -\frac{1}{3}$

• Equation: $y + 4 = -\frac{1}{3}(x - 1)$

Distance and Midpoint Formulas

Distance Formula

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

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

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

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

Midpoint Formula

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

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

Example: Find the midpoint between $(2, 5)$ and $(6, 9)$.

• $M = \left( \frac{2 + 6}{2}, \frac{5 + 9}{2} \right) = (4, 7)$

Equations of Circles

Standard Form of a Circle

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

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

• Center: $(h, k)$

• Radius: $r$

Example: Write the equation of a circle with center $(3, -2)$ and radius $5$.

• Equation: $(x - 3)^2 + (y + 2)^2 = 25$

Finding the Center and Radius from General Form

The general form of a circle is $x^2 + y^2 + Dx + Ey + F = 0$. To convert to standard form, complete the square for both $x$ and $y$ terms.

Example: Convert $x^2 + y^2 - 4x + 6y - 12 = 0$ to standard form.

• Group $x$ and $y$ terms: $(x^2 - 4x) + (y^2 + 6y) = 12$

• Complete the square: $(x - 2)^2 - 4 + (y + 3)^2 - 9 = 12$

• $(x - 2)^2 + (y + 3)^2 = 25$

• Center: $(2, -3)$, Radius: $5$

Applications and Problem Solving

Finding Equations from Graphs

• Identify the center and radius from the graph.

• Write the equation in standard form using the identified values.

Example: If a circle on a graph has center $(-1, 4)$ and radius $3$, the equation is$(x + 1)^2 + (y - 4)^2 = 9$.

Intersection Points

• To find where a line or curve intersects the axes, set $x = 0$ for the y-intercept and $y = 0$ for the x-intercept.

Example: For $y = 2x - 3$, the y-intercept is at $(0, -3)$ and the x-intercept is at $(1.5, 0)$.

Summary Table: Key Formulas

Concept	Formula	Description
Slope	$m = \frac{y_2 - y_1}{x_2 - x_1}$	Rate of change between two points
Distance	$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$	Length between two points
Midpoint	$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$	Point halfway between two points
Circle (Standard Form)	$(x - h)^2 + (y - k)^2 = r^2$	Equation of a circle
Line (Slope-Intercept)	$y = mx + b$	Equation of a line

Additional info: These notes cover essential Precalculus topics including equations of lines, circles, and their applications in coordinate geometry. Mastery of these concepts is foundational for further study in analytic geometry and calculus.