📐 Coordinate Geometry

Master coordinate geometry fundamentals including lines, circles, and basic conic sections. Essential for both AMC 10 and AMC 12.

🎯 Key Ideas

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

Circle Equation: $(x-h)^2 + (y-k)^2 = r^2$ with center $(h,k)$ and radius $r$

Line Equations: Various forms for different applications

📏 Distance and Midpoint

Distance Formula

Distance between points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$: $$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$

Midpoint Formula

Midpoint of segment from $P_1(x_1, y_1)$ to $P_2(x_2, y_2)$: $$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$$

Example: Find distance and midpoint between $(2,3)$ and $(-1,7)$

Solution:

  • Distance: $d = \sqrt{(-1-2)^2 + (7-3)^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9+16} = 5$
  • Midpoint: $\left(\frac{2+(-1)}{2}, \frac{3+7}{2}\right) = \left(\frac{1}{2}, 5\right)$

📈 Lines

Slope Formula

Slope of line through points $(x_1, y_1)$ and $(x_2, y_2)$: $$m = \frac{y_2-y_1}{x_2-x_1}$$

Line Equations

  • Point-slope: $y - y_1 = m(x - x_1)$
  • Slope-intercept: $y = mx + b$
  • Standard form: $Ax + By + C = 0$
  • Two-point form: $\frac{y-y_1}{y_2-y_1} = \frac{x-x_1}{x_2-x_1}$

Parallel and Perpendicular Lines

  • Parallel: Same slope ($m_1 = m_2$)
  • Perpendicular: Negative reciprocal slopes ($m_1 \cdot m_2 = -1$)

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

Solution:

  1. Slope of given line: $m_1 = 3$
  2. Slope of perpendicular: $m_2 = -\frac{1}{3}$
  3. Point-slope form: $y - 2 = -\frac{1}{3}(x - 1)$
  4. Simplify: $y = -\frac{1}{3}x + \frac{1}{3} + 2 = -\frac{1}{3}x + \frac{7}{3}$

⭕ Circles

Standard Form

Circle with center $(h,k)$ and radius $r$: $$(x-h)^2 + (y-k)^2 = r^2$$

General Form

$$x^2 + y^2 + Dx + Ey + F = 0$$

To convert to standard form, complete the square.

Example: Find center and radius of $x^2 + y^2 - 4x + 6y - 3 = 0$

Solution:

  1. Group terms: $(x^2 - 4x) + (y^2 + 6y) = 3$
  2. Complete squares:
    • $x^2 - 4x = (x-2)^2 - 4$
    • $y^2 + 6y = (y+3)^2 - 9$
  3. Substitute: $(x-2)^2 - 4 + (y+3)^2 - 9 = 3$
  4. Simplify: $(x-2)^2 + (y+3)^2 = 16$
  5. Center: $(2,-3)$, Radius: $4$

🔺 Area Formulas

Triangle Area

Area of triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$: $$A = \frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$$

Shoelace Formula

$$A = \frac{1}{2}\left|\sum_{i=1}^n (x_i y_{i+1} - x_{i+1} y_i)\right|$$ where $(x_{n+1}, y_{n+1}) = (x_1, y_1)$.

Example: Find area of triangle with vertices $(0,0)$, $(3,4)$, $(6,0)$

Solution:

  • $A = \frac{1}{2}|0(4-0) + 3(0-0) + 6(0-4)| = \frac{1}{2}|0 + 0 - 24| = \frac{1}{2} \cdot 24 = 12$

🎯 Basic Conic Sections (AMC 12 Light)

Parabola

  • Standard form: $y = ax^2 + bx + c$ or $x = ay^2 + by + c$
  • Vertex form: $y = a(x-h)^2 + k$ with vertex $(h,k)$

Ellipse

  • Standard form: $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$
  • Center: $(h,k)$
  • Major axis: $2a$ (longer axis)
  • Minor axis: $2b$ (shorter axis)

Hyperbola

  • Standard form: $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$
  • Center: $(h,k)$
  • Asymptotes: $y - k = \pm\frac{b}{a}(x - h)$

Example: Identify conic section $x^2 + 4y^2 - 2x + 8y + 1 = 0$

Solution:

  1. Complete squares:
    • $x^2 - 2x = (x-1)^2 - 1$
    • $4y^2 + 8y = 4(y^2 + 2y) = 4((y+1)^2 - 1) = 4(y+1)^2 - 4$
  2. Substitute: $(x-1)^2 - 1 + 4(y+1)^2 - 4 + 1 = 0$
  3. Simplify: $(x-1)^2 + 4(y+1)^2 = 4$
  4. Divide by 4: $\frac{(x-1)^2}{4} + \frac{(y+1)^2}{1} = 1$
  5. Answer: Ellipse with center $(1,-1)$, major axis 4, minor axis 2

🎯 AMC-Style Worked Example

Problem: Find the area of the triangle formed by the intersection of the lines $y = 2x + 1$, $y = -x + 4$, and $y = 3$.

Solution:

  1. Find intersection points:
    • Line 1 & 2: $2x + 1 = -x + 4$ → $3x = 3$ → $x = 1$ → $y = 3$ → $(1,3)$
    • Line 1 & 3: $2x + 1 = 3$ → $2x = 2$ → $x = 1$ → $(1,3)$
    • Line 2 & 3: $-x + 4 = 3$ → $x = 1$ → $(1,3)$
  2. Wait, all three lines pass through $(1,3)$!
  3. This means the three lines are concurrent, not forming a triangle
  4. Answer: Area = 0 (no triangle formed)

🔍 Common Traps & Fixes

Trap: Wrong slope calculation

Fix: $m = \frac{y_2-y_1}{x_2-x_1}$, not $\frac{x_2-x_1}{y_2-y_1}$.

Trap: Forgetting absolute value in area formula

Fix: Area is always positive, so use absolute value in triangle area formula.

Trap: Wrong circle center/radius

Fix: In $(x-h)^2 + (y-k)^2 = r^2$, center is $(h,k)$, not $(-h,-k)$.

Trap: Confusing conic section types

Fix: Check coefficients: $x^2$ and $y^2$ both positive = ellipse; opposite signs = hyperbola; one missing = parabola.

📋 Quick Reference

Distance and Midpoint

  • Distance: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
  • Midpoint: $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

Lines

  • Slope: $m = \frac{y_2-y_1}{x_2-x_1}$
  • Point-slope: $y - y_1 = m(x - x_1)$
  • Perpendicular: $m_1 \cdot m_2 = -1$

Circles

  • Standard: $(x-h)^2 + (y-k)^2 = r^2$
  • Center: $(h,k)$, Radius: $r$

Area

  • Triangle: $A = \frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$

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