Chapter 13: Coordinate Geometry
Learning Objectives
- Apply the distance and midpoint formulas to find lengths and midpoints in the coordinate plane
- Use slope to identify parallel and perpendicular lines
- Write equations of lines in slope-intercept, point-slope, and standard form
- Write and convert equations of circles; complete the square to find center and radius
- Apply the section formula to partition a directed segment in a given ratio
- Use coordinate geometry to prove geometric theorems
- Perform and describe translations, reflections, rotations, and dilations in the coordinate plane
13.1 Distance, Midpoint, and Slope
Coordinate geometry connects algebra and geometry by placing geometric figures in a coordinate plane. The three foundational tools are the distance formula, the midpoint formula, and slope.
Theorem: Distance, Midpoint, and Slope Formulas
- Distance: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ (derived from the Pythagorean Theorem)
- Midpoint: $M = \left(\dfrac{x_1 + x_2}{2},\ \dfrac{y_1 + y_2}{2}\right)$
- Slope: $m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{\text{rise}}{\text{run}}$
- Parallel lines: equal slopes ($m_1 = m_2$)
- Perpendicular lines: negative reciprocal slopes ($m_1 \cdot m_2 = -1$)
Example 13.1 — Distance, Midpoint, and Slope
Given $A = (2, 3)$ and $B = (8, 11)$, find $AB$, the midpoint $M$, and the slope of $\overline{AB}$.
Distance: $d = \sqrt{(8-2)^2 + (11-3)^2} = \sqrt{36 + 64} = \sqrt{100} = \mathbf{10}$
Midpoint: $M = \left(\dfrac{2+8}{2},\ \dfrac{3+11}{2}\right) = \mathbf{(5, 7)}$
Slope: $m = \dfrac{11-3}{8-2} = \dfrac{8}{6} = \mathbf{\dfrac{4}{3}}$
Example 13.2 — Parallel and Perpendicular Lines
A line through $A = (0, 0)$ has slope $3$: equation $y = 3x$.
A line perpendicular to it through $A$ has slope $m = -\dfrac{1}{3}$: equation $y = -\dfrac{1}{3}x$.
Any line parallel to $y = 3x$ also has slope $3$, but a different $y$-intercept.
Find the perimeter of the triangle with vertices $P = (0, 0)$, $Q = (6, 0)$, $R = (3, 4)$.
Show Answer
$QR = \sqrt{(6-3)^2 + (0-4)^2} = \sqrt{9 + 16} = 5$
$PR = \sqrt{(0-3)^2 + (0-4)^2} = \sqrt{9 + 16} = 5$
Perimeter $= 6 + 5 + 5 = \mathbf{16}$
Right triangle showing the distance formula derivation — the hypotenuse is the distance, horizontal and vertical legs are the coordinate differences.
Figure 13.1 — Distance Formula from the Pythagorean Theorem
13.2 Equations of Lines and Circles
Every non-vertical line in the plane has a slope. Three standard forms of line equations allow us to write and interpret lines efficiently.
Forms of Line Equations
- Slope-intercept: $y = mx + b$ (slope $m$, $y$-intercept $b$)
- Point-slope: $y - y_1 = m(x - x_1)$ (slope $m$, point $(x_1, y_1)$)
- Standard form: $Ax + By = C$ (integers, $A \geq 0$)
A circle with center $(h, k)$ and radius $r$ is the set of all points equidistant from $(h, k)$. Applying the distance formula gives the standard equation.
Theorem: Standard Form of a Circle
$(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.
The general form $x^2 + y^2 + Dx + Ey + F = 0$ can be converted by completing the square on both $x$ and $y$ terms.
Example 13.3 — Equation of a Circle
Write the equation of the circle with center $(3, -2)$ and radius $5$.
$(x - 3)^2 + (y + 2)^2 = 25$
Example 13.4 — Completing the Square
Convert $x^2 + y^2 - 6x + 4y - 3 = 0$ to standard form. Find the center and radius.
Group: $(x^2 - 6x) + (y^2 + 4y) = 3$
Complete the square: $(x^2 - 6x + 9) + (y^2 + 4y + 4) = 3 + 9 + 4$
$(x - 3)^2 + (y + 2)^2 = 16$
Center: $(3, -2)$, $\quad$ Radius: $r = 4$
Example 13.5 — Line Parallel to a Given Line
Find the equation of the line through $(4, 7)$ parallel to $y = 2x - 3$.
Parallel lines have equal slopes: $m = 2$.
Point-slope: $y - 7 = 2(x - 4) \Rightarrow y = 2x - 8 + 7 \Rightarrow \mathbf{y = 2x - 1}$
Find the equation of the line through $(1, 5)$ perpendicular to $y = 3x + 2$.
Show Answer
$y - 5 = -\dfrac{1}{3}(x - 1) \Rightarrow y = -\dfrac{1}{3}x + \dfrac{1}{3} + 5 = -\dfrac{1}{3}x + \dfrac{16}{3}$
Circle $(x-3)^2+(y-2)^2=9$ with center marked and radius drawn — axes visible for reference.
Figure 13.2 — Circle in Standard Form: Center and Radius
13.3 Partitioning a Segment and Coordinate Proofs
The section formula (also called the partition formula) finds the point $P$ that divides segment $\overline{AB}$ in a given ratio from $A$.
Theorem: Section Formula (Partitioning)
The point $P$ that divides $\overline{AB}$ with $A = (x_1, y_1)$ and $B = (x_2, y_2)$ in ratio $m:n$ from $A$ is:
$$P = \left(\frac{nx_1 + mx_2}{m + n},\ \frac{ny_1 + my_2}{m + n}\right)$$
Equivalently: $P = A + \dfrac{m}{m+n}(B - A)$, i.e., move $\dfrac{m}{m+n}$ of the way from $A$ to $B$.
Coordinate geometry proofs use the distance and midpoint formulas to verify geometric properties with algebra:
- Prove two segments are congruent: show their lengths are equal (distance formula)
- Prove two lines are parallel: show their slopes are equal
- Prove a point is a midpoint: show it equals the midpoint formula result
- Prove diagonals bisect each other: show they share the same midpoint
Example 13.6 — Partitioning a Segment
Divide $\overline{AB}$ with $A = (0, 0)$ and $B = (12, 6)$ in ratio $1:3$ from $A$.
$P = \left(\dfrac{1 \cdot 12}{1+3},\ \dfrac{1 \cdot 6}{1+3}\right) = \left(\dfrac{12}{4},\ \dfrac{6}{4}\right) = \mathbf{(3,\ 1.5)}$
Example 13.7 — Coordinate Proof: Diagonals of a Square Are Congruent
Given: Square $ABCD$ with $A = (0,0)$, $B = (a, 0)$, $C = (a, a)$, $D = (0, a)$. Prove $AC = BD$.
$AC = \sqrt{(a-0)^2 + (a-0)^2} = \sqrt{2a^2} = a\sqrt{2}$
$BD = \sqrt{(0-a)^2 + (a-0)^2} = \sqrt{a^2 + a^2} = a\sqrt{2}$
$AC = BD = a\sqrt{2}$ $\checkmark$ — the diagonals are congruent. $\square$
Divide segment from $P = (2, 4)$ to $Q = (10, 12)$ in ratio $3:1$ from $P$. Find the point.
Show Answer
$x = 2 + \dfrac{3}{4}(10 - 2) = 2 + 6 = 8$
$y = 4 + \dfrac{3}{4}(12 - 4) = 4 + 6 = 10$
Point: $(8, 10)$
13.4 Transformations in the Coordinate Plane
A transformation is a mapping of each point in the plane to another point. Rigid transformations (isometries) preserve distances and angle measures. Dilations preserve shape but change size.
Transformation Rules Table
| Transformation | Rule: $(x, y) \to$ | Preserves |
|---|---|---|
| Translation by $(a, b)$ | $(x+a,\ y+b)$ | Size and shape (rigid) |
| Reflection over $x$-axis | $(x,\ -y)$ | Size and shape (rigid) |
| Reflection over $y$-axis | $(-x,\ y)$ | Size and shape (rigid) |
| Reflection over $y = x$ | $(y,\ x)$ | Size and shape (rigid) |
| Reflection over origin | $(-x,\ -y)$ | Size and shape (rigid) |
| Rotation 90° CCW | $(-y,\ x)$ | Size and shape (rigid) |
| Rotation 180° | $(-x,\ -y)$ | Size and shape (rigid) |
| Rotation 270° CCW | $(y,\ -x)$ | Size and shape (rigid) |
| Dilation by factor $k$ | $(kx,\ ky)$ | Shape only (similarity) |
Example 13.8 — Reflection and Rotation
$\triangle PQR$ has $P = (2,3)$, $Q = (5,3)$, $R = (5,7)$.
Reflection over $y$-axis: $P' = (-2,3)$, $Q' = (-5,3)$, $R' = (-5,7)$
Rotation 90° CCW from original: Rule $(-y, x)$:
$P'' = (-3, 2)$, $Q'' = (-3, 5)$, $R'' = (-7, 5)$
Example 13.9 — Dilation
Dilate $\triangle PQR$ with scale factor $k = 2$, center at origin.
Rule: $(x, y) \to (2x, 2y)$
$P' = (4, 6)$, $Q' = (10, 6)$, $R' = (10, 14)$
The image is similar to the original with all sides doubled.
Reflect $A = (3, -4)$ over the line $y = x$. What is $A'$?
Show Answer
$A' = (-4, 3)$
AP Tip: Rigid transformations (translation, reflection, rotation) produce congruent figures — same size and shape. Dilation produces a similar figure — same shape, different size. Only rigid transformations preserve congruence; dilations (unless $k = 1$) do not.
Original triangle ABC (purple) and its reflection over the y-axis (blue) — the y-axis acts as a line of symmetry.
Figure 13.3 — Reflection over the Y-Axis
Practice Problems
Find the distance between $(-3, 2)$ and $(5, -4)$, and find the midpoint.
Show Solution
Midpoint: $M = \left(\dfrac{-3+5}{2},\ \dfrac{2+(-4)}{2}\right) = \mathbf{(1, -1)}$
Prove that $A = (0,0)$, $B = (6,0)$, $C = (6,4)$, $D = (0,4)$ is a rectangle. Use slopes and distances.
Show Solution
Opposite sides equal: $AB = CD = 6$; $BC = DA = 4$.
$ABCD$ has four right angles and opposite sides equal. $\Rightarrow$ Rectangle. $\square$
Find the equation of the circle with center $(-2, 3)$ and passing through $(2, 6)$.
Show Solution
Equation: $(x+2)^2 + (y-3)^2 = 25$
Convert $x^2 + y^2 + 8x - 10y + 5 = 0$ to standard form. Find the center and radius.
Show Solution
$(x+4)^2 + (y-5)^2 = 36$
Center: $(-4, 5)$, $\quad$ Radius: $r = 6$
Find the point that divides the segment from $A = (-4, 2)$ to $B = (8, 14)$ in ratio $2:1$ from $A$.
Show Solution
$x = -4 + \dfrac{2}{3}(8-(-4)) = -4 + \dfrac{2}{3}(12) = -4 + 8 = 4$
$y = 2 + \dfrac{2}{3}(14-2) = 2 + 8 = 10$
Point: $(4, 10)$
Write the transformation rule for reflecting over the $x$-axis. Apply it to $A = (3, 4)$. What is $A'$?
Show Solution
$A' = (3, -4)$
The $x$-coordinate stays the same; the $y$-coordinate changes sign.
Dilate $\triangle JKL$ with $J = (1,2)$, $K = (3,0)$, $L = (2,4)$ by factor $k = 3$ from the origin. Find $J'$, $K'$, $L'$ and the ratio of perimeters.
Show Solution
For a dilation with scale factor $k$, all lengths scale by $k$.
Ratio of perimeters $\triangle J'K'L' : \triangle JKL = 3 : 1$
AP Problem: Use coordinate geometry to prove that the diagonals of a parallelogram bisect each other. Place the parallelogram at $A = (0,0)$, $B = (a,0)$, $C = (a+b,c)$, $D = (b,c)$.
Show Solution
Midpoint $= \left(\dfrac{a+b}{2},\ \dfrac{c}{2}\right)$
Diagonal $BD$: from $(a, 0)$ to $(b, c)$
Midpoint $= \left(\dfrac{a+b}{2},\ \dfrac{c}{2}\right)$
Both diagonals share the same midpoint $\Rightarrow$ the diagonals bisect each other. $\square$
📋 Chapter Summary
Key Formulas
$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ — derived from the Pythagorean theorem.
$M = \left(\dfrac{x_1+x_2}{2},\, \dfrac{y_1+y_2}{2}\right)$ — the average of coordinates.
$m = \dfrac{y_2-y_1}{x_2-x_1} = \dfrac{\text{rise}}{\text{run}}$. Parallel lines: equal slopes. Perpendicular: slopes are negative reciprocals.
Slope-intercept: $y = mx + b$. Point-slope: $y - y_1 = m(x - x_1)$. Standard: $Ax + By = C$.
Coordinate Proof Strategy
- Place figure conveniently — put one vertex at origin, align sides with axes
- Use general coordinates — label vertices with variables like $(a, 0)$, $(0, b)$
- Apply formulas — distance, midpoint, slope as needed
- State conclusion — connect algebraic result to geometric property