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AMC 10 & 12

Coordinate Geometry

Turning a geometry problem into algebra — lines, circles, and the shoelace formula for polygon area from vertices alone.

Lines

m=y2−y1x2−x1(slope)m = \dfrac{y_2-y_1}{x_2-x_1} \qquad\text{(slope)}
y−y1=m(x−x1)(point-slope form)y - y_1 = m(x-x_1) \qquad\text{(point-slope form)}

Parallel lines share the same slope; perpendicular lines have slopes that are negative reciprocals of each other (m1m2=−1m_1 m_2 = -1).

Distance and midpoint

d=(x2−x1)2+(y2−y1)2,midpoint=(x1+x22,y1+y22)d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, \qquad \text{midpoint} = \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)

Circles

Standard form directly gives the center and radius:

(x−h)2+(y−k)2=r2center (h,k), radius r(x-h)^2+(y-k)^2=r^2 \qquad\text{center } (h,k),\ \text{radius } r

General form requires completing the square to extract the same information:

x2+y2+Dx+Ey+F=0center (−D2,−E2)x^2+y^2+Dx+Ey+F=0 \qquad\text{center } \left(-\dfrac{D}{2}, -\dfrac{E}{2}\right)

The shoelace formula

Given a polygon's vertices in order (either clockwise or counterclockwise), the shoelace formula computes its area directly — no need to decompose it into triangles by hand:

Area=12∣∑i(xiyi+1−xi+1yi)∣\text{Area} = \dfrac12\left| \sum_{i} (x_i y_{i+1} - x_{i+1} y_i) \right|

For a triangle with vertices (x1,y1),(x2,y2),(x3,y3)(x_1,y_1), (x_2,y_2), (x_3,y_3), this expands to:

Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\text{Area} = \dfrac12\left| x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \right|

Worked example

Problem: Find the area of the triangle with vertices (1,1)(1,1), (4,5)(4,5), and (7,2)(7,2).

Solution: Apply the shoelace formula:

Area=12∣1(5−2)+4(2−1)+7(1−5)∣=12∣3+4−28∣=12(21)=212\text{Area} = \dfrac12\left|1(5-2)+4(2-1)+7(1-5)\right| = \dfrac12\left|3+4-28\right| = \dfrac12(21) = \dfrac{21}{2}

Practice problems

1.Find the slope of the line through (2,3)(2,3) and (6,11)(6,11).

2.Find the center and radius of x2+y2−6x+4y−12=0x^2+y^2-6x+4y-12=0.

3.Find the distance between (−1,2)(-1,2) and (3,−1)(3,-1).

4.Use the shoelace formula to find the area of the quadrilateral with vertices (0,0),(4,0),(4,3),(0,3)(0,0), (4,0), (4,3), (0,3) (in order).

Tangency, Incircles & ExcirclesTransformations