6.1 Coordinate Geometry: Lines, Distance, Midpoint, and Circles

6.1 Coordinate Geometry: Lines, Distance, Midpoint, and Circles

Analytic geometry provides the foundational bridge between algebraic equations and geometric figures on the Cartesian coordinate plane $\mathbb{R}^2$. A mastery of fundamental geometric measurements—such as distance, midpoints, slopes, and linear relationships—is essential for evaluating complex geometric systems and second-degree curves on the CLEP Precalculus examination.

---

1. Distance and Midpoint Formulas in $\mathbb{R}^2$

The Distance Formula

The distance $d$ between any two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ in the Cartesian plane is derived directly from the Pythagorean theorem ($a^2 + b^2 = c^2$). By constructing a right triangle with horizontal leg $\Delta x = |x_2 - x_1|$ and vertical leg $\Delta y = |y_2 - y_1|$, the hypotenuse represents the Euclidean distance:

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

Because the differences are squared, the order in which the points are subtracted does not affect the calculation: $(x_2 - x_1)^2 = (x_1 - x_2)^2$.

The Midpoint Formula

The midpoint $M$ of a line segment joining endpoints $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ is the point whose coordinates are the arithmetic averages of the respective coordinates of the endpoints:

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

Worked Example 1: Segment Analysis & Right-Triangle Verification

Given three points $P_1(-3, 7)$, $P_2(5, 1)$, and $P_3(5, 7)$:

1. Compute the length of segment $P_1 P_2$ and find its midpoint $M$.
2. Verify that triangle $\triangle P_1 P_3 P_2$ is a right-angled triangle.

Solution:

1. Calculate the distance $d(P_1, P_2)$: $d = \sqrt{(5 - (-3))^2 + (1 - 7)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$ Calculate the midpoint $M$ of segment $P_1 P_2$: $M = \left(\frac{-3 + 5}{2}, \frac{7 + 1}{2}\right) = \left(\frac{2}{2}, \frac{8}{2}\right) = (1, 4)$
2. Calculate the remaining side lengths of $\triangle P_1 P_3 P_2$: $d(P_1, P_3) = \sqrt{(5 - (-3))^2 + (7 - 7)^2} = \sqrt{8^2 + 0^2} = 8$ $d(P_3, P_2) = \sqrt{(5 - 5)^2 + (1 - 7)^2} = \sqrt{0^2 + (-6)^2} = 6$ Check the Pythagorean identity $a^2 + b^2 = c^2$: $[d(P_1, P_3)]^2 + [d(P_3, P_2)]^2 = 8^2 + 6^2 = 64 + 36 = 100 = 10^2 = [d(P_1, P_2)]^2$ Because the side lengths satisfy $8^2 + 6^2 = 10^2$, the triangle is a right triangle with its right angle located at $P_3(5, 7)$.

---

2. Equations of Lines, Parallel & Perpendicular Slopes, and Point-to-Line Distance

Linear Forms

A line in the Cartesian plane can be expressed in several algebraic forms:

• Slope-Intercept Form: $y = mx + b$, where $m$ is the slope and $b$ is the $y$-intercept $(0, b)$.
• Point-Slope Form: $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is a known point.
• General (Standard) Form: $Ax + By + C = 0$, where $A$, $B$, and $C$ are integers, and $A \ge 0$.

Slope Relationships

For non-vertical lines $L_1$ and $L_2$ with slopes $m_1$ and $m_2$:

• Parallel Lines ($L_1 \parallel L_2$): Slopes are equal: $m_1 = m_2$.
• Perpendicular Lines ($L_1 \perp L_2$): Slopes are negative reciprocals: $m_1 \cdot m_2 = -1 \implies m_2 = -\frac{1}{m_1}$.

Point-to-Line Distance Formula

The shortest distance $d$ from a point $P(x_1, y_1)$ to a line defined by $Ax + By + C = 0$ is measured along a perpendicular segment dropped from the point to the line:

$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Worked Example 2: Perpendicular Line Equation & Distance

Find the general equation of the line $L_2$ passing through $P(2, -5)$ that is perpendicular to $L_1: 3x - 4y + 12 = 0$. Then calculate the exact perpendicular distance from $P(2, -5)$ to $L_1$.

Solution:

1. Find the slope of $L_1$ by converting to slope-intercept form: $-4y = -3x - 12 \implies y = \frac{3}{4}x + 3 \implies m_1 = \frac{3}{4}$
2. Determine the slope $m_2$ of perpendicular line $L_2$: $m_2 = -\frac{1}{m_1} = -\frac{4}{3}$
3. Use point-slope form with $P(2, -5)$: $y - (-5) = -\frac{4}{3}(x - 2) \implies y + 5 = -\frac{4}{3}x + \frac{8}{3}$ Multiply through by $3$ to obtain general form: $3y + 15 = -4x + 8 \implies 4x + 3y + 7 = 0$
4. Calculate the perpendicular distance from $P(2, -5)$ to $L_1: 3x - 4y + 12 = 0$: $d = \frac{|3(2) - 4(-5) + 12|}{\sqrt{3^2 + (-4)^2}} = \frac{|6 + 20 + 12|}{\sqrt{9 + 16}} = \frac{|38|}{\sqrt{25}} = \frac{38}{5} = 7.6$

---

3. Standard and General Equations of Circles

Locus Definition and Standard Form

A circle is the set (locus) of all points $P(x, y)$ in a plane that are at a fixed positive distance $r$ (the radius) from a fixed point $C(h, k)$ (the center). Applying the distance formula yields the standard equation of a circle:

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

If the center is at the origin $(0, 0)$, the equation simplifies to $x^2 + y^2 = r^2$.

General Form and Completing the Square

Expanding the standard equation yields the general form of a circle:

$Ax^2 + Ay^2 + Dx + Ey + F = 0 \quad (A \neq 0)$

To identify the center $(h, k)$ and radius $r$ from general form, divide by $A$ and complete the square for both $x$ and $y$ terms.

Worked Example 3: Completing the Square for a Circle

Determine the center, radius, and area of the circle given by the general equation:

$2x^2 + 2y^2 - 12x + 16y - 14 = 0$

Solution:

1. Divide every term by $A = 2$: $x^2 + y^2 - 6x + 8y - 7 = 0$
2. Rearrange terms by grouping variables and isolating the constant on the right side: $(x^2 - 6x) + (y^2 + 8y) = 7$
3. Complete the square for $x$ by adding $\left(\frac{-6}{2}\right)^2 = 9$, and for $y$ by adding $\left(\frac{8}{2}\right)^2 = 16$. Add both values to the right side: $(x^2 - 6x + 9) + (y^2 + 8y + 16) = 7 + 9 + 16$
4. Factor the perfect square trinomials into standard form: $(x - 3)^2 + (y + 4)^2 = 32$
5. Compare with $(x - h)^2 + (y - k)^2 = r^2$:
   • Center: $(h, k) = (3, -4)$
   • Radius: $r = \sqrt{32} = 4\sqrt{2}$
   • Area: $A = \pi r^2 = \pi (32) = 32\pi$ square units.

---

4. CLEP Exam Traps & Common Errors