COORDINATE GEOMETRY

Overview

Coordinate geometry translates geometric objects — points, lines, triangles, polygons — into algebra on the $xy$-plane. Many AMC 10 problems that look impossible to approach purely with classical geometry become routine once you assign coordinates wisely and apply a small set of formulas. The key skill is choosing coordinates that simplify the arithmetic: place one vertex at the origin, align a side along the $x$-axis, or center a figure at the origin when the problem has symmetry.

The AMC 10 draws on four main coordinate topics: distance and midpoints, slope and line equations (especially parallel/perpendicular relationships), area by the shoelace formula, and reflections. A fifth topic — the distance from a point to a line — appears less often but is worth knowing, especially for problems about altitudes of triangles or minimum distances.

Understanding these tools deeply means knowing not just the formula but also when to reach for each one. Distance answers "how far?", midpoint answers "where is the middle?", slope answers "how steep or which direction?", and shoelace turns a list of vertices into an area with no geometry needed.

Key facts

• Distance formula: The distance between points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$ Use whenever two coordinates are given and a length is needed.

• Midpoint formula: The midpoint of the segment from $(x_1, y_1)$ to $(x_2, y_2)$ is $M = \left(\frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2}\right).$ Use to find a center, bisect a segment, or exploit symmetry.

• Slope: The slope of a line through $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}.$ Slope measures rise over run; a horizontal line has slope $0$; a vertical line has undefined slope.

• Parallel lines: Two lines are parallel if and only if they have equal slopes ($m_1 = m_2$). Or both are vertical.

• Perpendicular lines: Two non-vertical lines are perpendicular if and only if their slopes are negative reciprocals: $m_1 \cdot m_2 = -1$. Flip and negate one slope to get the other.

• Slope-intercept form: $y = mx + b$, where $m$ is the slope and $b$ is the $y$-intercept.

• Point-slope form: $y - y_1 = m(x - x_1)$, the line of slope $m$ through $(x_1, y_1)$. Fastest way to write a line equation from a point and a slope.

• Standard (general) form: $ax + by + c = 0$. The slope is $-a/b$ (for $b \neq 0$).

• Shoelace formula (area of polygon): For a polygon with vertices $(x_1,y_1), (x_2,y_2), \ldots, (x_n,y_n)$ listed in order (clockwise or counter-clockwise), $\text{Area} = \frac{1}{2}\left|\sum_{i=1}^{n}\bigl(x_i y_{i+1} - x_{i+1} y_i\bigr)\right|,$ where indices wrap around so $(x_{n+1}, y_{n+1}) = (x_1, y_1)$. Works for any polygon; no base-height needed.

• Reflection over $y = x$: The reflection of $(a, b)$ over the line $y = x$ is $(b, a)$. Swap the coordinates.

• Reflection over a vertical line $x = h$: The reflection of $(a, b)$ is $(2h - a,\; b)$. The $y$-coordinate is unchanged; the $x$-coordinate is "mirrored" around $h$.

• Reflection over a horizontal line $y = k$: The reflection of $(a, b)$ is $(a,\; 2k - b)$.

• Rotation by $90°$ counter-clockwise about the origin: $(a, b) \mapsto (-b, a)$.

• Rotation by $180°$ about the origin: $(a, b) \mapsto (-a, -b)$.

• Distance from a point to a line: The distance from point $(x_0, y_0)$ to the line $ax + by + c = 0$ is $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}.$ The absolute value and the $\sqrt{a^2+b^2}$ denominator are both easy to forget — write them first.

Worked example 1

Points $A = (1, 3)$ and $B = (7, 11)$. Find the length of $\overline{AB}$, the midpoint $M$, and the equation of the perpendicular bisector of $\overline{AB}$.

Length: $AB = \sqrt{(7-1)^2 + (11-3)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.

Midpoint: $M = \left(\dfrac{1+7}{2}, \dfrac{3+11}{2}\right) = (4, 7)$.

Perpendicular bisector: The slope of $\overline{AB}$ is $\dfrac{11-3}{7-1} = \dfrac{8}{6} = \dfrac{4}{3}$. The perpendicular bisector passes through $M = (4,7)$ with slope $-\dfrac{3}{4}$ (negative reciprocal).

Using point-slope form: $y - 7 = -\frac{3}{4}(x - 4) \implies y = -\frac{3}{4}x + 10.$

Why does this work? Every point on the perpendicular bisector of $\overline{AB}$ is equidistant from $A$ and $B$. The two conditions "passes through the midpoint" and "slope is the negative reciprocal" encode exactly that requirement.

Worked example 2

A triangle has vertices $P = (0, 0)$, $Q = (8, 0)$, $R = (3, 6)$. Find its area using the shoelace formula, and find the equation of the altitude from $R$ to $\overline{PQ}$.

Area by shoelace: List the vertices in order and apply the formula:

$\text{Area} = \frac{1}{2}|x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)|$ $= \frac{1}{2}|0(0-6) + 8(6-0) + 3(0-0)| = \frac{1}{2}|0 + 48 + 0| = 24.$

Altitude from $R$: $\overline{PQ}$ lies on the $x$-axis and has slope $0$. The altitude from $R$ is perpendicular to $\overline{PQ}$, so it is a vertical line passing through $R = (3, 6)$: the equation is $x = 3$.

Check: The altitude has length equal to the $y$-coordinate of $R$, which is $6$. Using $\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 8 \cdot 6 = 24$. This matches the shoelace result.

Common traps

• Forgetting the $\frac{1}{2}$ in the shoelace formula. The "shoelace sum" before dividing is twice the area. Forgetting the factor of $\frac{1}{2}$ is the single most common shoelace error.
• Using slope = run/rise instead of rise/run. Slope is always $\Delta y / \Delta x$, not $\Delta x / \Delta y$. When in doubt, write it out as (change in output)/(change in input).
• Perpendicular slope: flipping without negating (or negating without flipping). If $m = 2/3$, the perpendicular slope is $-3/2$, not $3/2$ or $-2/3$. Both operations (reciprocal AND sign change) are required.
• Reflection formula sign errors. Reflecting $(a,b)$ over the line $x = h$ gives $x' = 2h - a$, not $2h + a$. Sketch a one-dimensional number line — the image is as far to the right of $h$ as $a$ is to the left.
• Plugging the wrong sign into the point-to-line distance. The formula requires the line in the form $ax + by + c = 0$. If you write $3x - 4y = 5$, rewrite it as $3x - 4y - 5 = 0$ before substituting; $c = -5$, not $+5$.