1.6 Coordinate Geometry: Lines, Distance, Midpoints & Conic Sections
Key Takeaways
- Calculate distance, midpoint, and slope in the Cartesian plane using explicit coordinate formulas.
- Derive line equations using slope-intercept, point-slope, and standard form, enforcing parallel/perpendicular slope relationships.
- Calculate the perpendicular distance from a point to a line using the standard distance formula.
- Identify conic sections (circles, parabolas, ellipses, hyperbolas) from general quadratic equations using the discriminant B^2 - 4AC.
Coordinate Geometry: Lines, Distance, Midpoints & Conic Sections
Coordinate geometry bridges algebra and plane geometry, representing one of the most heavily tested content domains on the Ateneo College Entrance Test (ACET) Mathematics subtest. Success on the ACET requires rapid execution of distance and midpoint formulas, linear slope derivations, perpendicular distance calculations from points to lines, Shoelace polygon area determinations, and structural identification of conic sections (circles, parabolas, ellipses, hyperbolas).
1. Fundamental Cartesian Formulas
Given two arbitrary points in the Cartesian plane $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$:
Primary Metric Formulas
- Distance Formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
- Midpoint Formula: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$
- Slope Formula: $m = \frac{y_2 - y_1}{x_2 - x_1} \quad (x_1 \neq x_2)$
- Section Formula (Internal Partitioning ratio $m:n$): $P = \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right)$
- Centroid of a Triangle $(x_1, y_1), (x_2, y_2), (x_3, y_3)$: $C = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$
2. Linear Equations, Parallelism & Perpendicular Distance
Linear Forms
- Slope-Intercept Form: $y = mx + b$ ($m$ is slope, $b$ is y-intercept)
- Point-Slope Form: $y - y_1 = m(x - x_1)$
- General Standard Form: $Ax + By + C = 0$ (where slope $m = -A/B$, y-intercept = $-C/B$)
Parallel & Perpendicular Line Relationships
- Parallel Lines ($L_1 \parallel L_2$): Slopes are equal ($m_1 = m_2$). Lines never intersect unless identical.
- Perpendicular Lines ($L_1 \perp L_2$): Slopes are negative reciprocals ($m_1 \cdot m_2 = -1 \implies m_2 = -\frac{1}{m_1}$).
Perpendicular Distance from Point $(x_0, y_0)$ to Line $Ax + By + C = 0$
The shortest distance from an external point $(x_0, y_0)$ to a straight line expressed in general form $Ax + By + C = 0$ is:
Shoelace Formula for Polygon Area
To calculate the area of a polygon with ordered vertices $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$:
3. Conic Sections: Canonical Equations & Structural Properties
A conic section is formed by the intersection of a plane and a double-napped cone. In general form without xy-rotation ($B=0$), conics follow:
Conic Section Taxonomy & Properties Table
| Conic Section | Standard Canonical Equation | Center / Vertex | Key Metric Identities & Foci |
|---|---|---|---|
| Circle | $(x-h)^2 + (y-k)^2 = r^2$ | Center $(h,k)$ | Radius $r = \sqrt{r^2}$, Constant distance from center |
| Parabola (Vertical) | $(x-h)^2 = 4p(y-k)$ | Vertex $(h,k)$ | Focus $(h, k+p)$, Directrix $y = k-p$, $p \neq 0$ |
| Parabola (Horizontal) | $(y-k)^2 = 4p(x-h)$ | Vertex $(h,k)$ | Focus $(h+p, k)$, Directrix $x = h-p$, $p \neq 0$ |
| Ellipse (Horizontal) | $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ | Center $(h,k)$ | $c^2 = a^2 - b^2 \quad (a > b)$, Foci $(h \pm c, k)$ |
| Hyperbola (Horizontal) | $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ | Center $(h,k)$ | $c^2 = a^2 + b^2$, Asymptotes $y-k = \pm \frac{b}{a}(x-h)$ |
Identifying Conics via Coefficients ($A x^2 + C y^2 + D x + E y + F = 0$)
- Parabola: $A = 0$ OR $C = 0$ (contains exactly one squared term).
- Circle: $A = C \neq 0$ (squared terms have identical non-zero coefficients with the same sign).
- Ellipse: $A \neq C$, but $A \cdot C > 0$ (squared terms have different positive coefficients of the same sign).
- Hyperbola: $A \cdot C < 0$ (squared terms have coefficients of opposite signs).
Worked Step-by-Step ACET Exam Problems
Problem 1: Perpendicular Distance from Point to Line
Find the exact perpendicular distance from point $P(3, -2)$ to the line $5x - 12y + 7 = 0$.
Step 1: Extract general line coefficients and point coordinates. $A = 5$, $B = -12$, $C = 7$, $x_0 = 3$, $y_0 = -2$.
Step 2: Substitute into point-to-line distance formula. $d = \frac{|A x_0 + B y_0 + C|}{\sqrt{A^2 + B^2}} = \frac{|5(3) + (-12)(-2) + 7|}{\sqrt{5^2 + (-12)^2}}$
Step 3: Evaluate absolute numerator and radical denominator.
- Numerator: $|15 + 24 + 7| = |46| = 46$
- Denominator: $\sqrt{25 + 144} = \sqrt{169} = 13$
- Final Distance: $d = \frac{46}{13} \approx 3.538$.
Problem 2: Completing the Square to Analyze a Circle Equation
Determine the center $(h, k)$ and radius $r$ of the circle defined by $x^2 + y^2 - 6x + 8y - 11 = 0$.
Step 1: Group x and y variables and isolate constant term. $(x^2 - 6x) + (y^2 + 8y) = 11$.
Step 2: Complete the square for both variable groups.
- For $x$: $\left(\frac{-6}{2}\right)^2 = (-3)^2 = 9$.
- For $y$: $\left(\frac{8}{2}\right)^2 = (4)^2 = 16$. Add $9 + 16 = 25$ to both sides of the equation: $(x^2 - 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16$.
Step 3: Factor squared terms into standard canonical form. $(x - 3)^2 + (y + 4)^2 = 36$.
Step 4: Read center coordinates and radius.
- Center $(h, k) = (3, -4)$.
- Radius $r = \sqrt{36} = 6$.
ACET Speed Tactics & Traps Summary
| Problem Type | Common Student Trap | ACET Speed Tactic |
|---|---|---|
| Perpendicular Lines | Using $-m$ instead of negative reciprocal $-1/m$ | Negative reciprocal flips numerator/denominator and sign |
| Circle Radius | Forgetting to add completed terms to right side | Add $(D/2A)^2 + (E/2C)^2$ to constant side immediately |
| Ellipse Foci Metric | Confusing $c^2 = a^2 - b^2$ with hyperbola $c^2 = a^2 + b^2$ | Ellipse subtracts ($a^2 - b^2$); Hyperbola adds ($a^2 + b^2$) |
| Midpoint Evaluation | Subtracting coordinates instead of taking average | Midpoint is coordinate average: add values and divide by 2 |
| Slope of Vertical Line | Setting vertical slope $m = 0$ instead of undefined | Horizontal line has $m=0$; Vertical line has undefined slope |
What is the equation of the line passing through point (2, -5) and perpendicular to the line 3x - 4y = 12?
Which conic section is represented by the second-degree equation 4x^2 - 9y^2 - 16x + 18y - 29 = 0?