Coordinate Geometry

What you need to know

• Distance, midpoint, slope. For $(x_1,y_1)$ and $(x_2,y_2)$: distance $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ (the Pythagorean Theorem), midpoint $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$, slope $\frac{y_2-y_1}{x_2-x_1}$. Perpendicular lines have slopes that multiply to $-1$ (when neither line is vertical). Example: from $(1,2)$ to $(4,6)$, the distance is $\sqrt{9+16} = 5$, the midpoint is $\left(\frac52, 4\right)$, the slope is $\frac43$, and a perpendicular line has slope $-\frac34$.
• Distance from a point to a line. From $(x_0,y_0)$ to the line $ax+by+c=0$, it is $\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$. A line is tangent to a circle exactly when this distance from the center equals the radius. Example: from $(0,0)$ to $3x+4y-10=0$ the distance is $\frac{10}{5} = 2$, so the line is tangent to the circle of radius $2$ centered at $(0,0)$.
• Circles. The circle with center $(h,k)$ and radius $r$ is $(x-h)^2+(y-k)^2=r^2$. Given $x^2+y^2+Dx+Ey+F=0$ instead, complete the square in $x$ and in $y$ to read off the center and radius. Example: $x^2+y^2-6x+4y-12=0$ becomes $(x-3)^2+(y+2)^2 = 25$: center $(3,-2)$, radius $5$.
• Reflections. Over the $x$-axis, $(x,y)$ goes to $(x,-y)$. Over $y=x$, swap to get $(y,x)$. Over $y=-x$, swap and change both signs to get $(-y,-x)$. Example: $(2,5)$ goes to $(2,-5)$, $(5,2)$, and $(-5,-2)$.
• Lattice points (points whose coordinates are both integers). If $a$ and $b$ are integers, the segment from $(0,0)$ to $(a,b)$ contains $\gcd(a,b)+1$ lattice points, counting both ends. Example: to $(6,9)$, $\gcd = 3$, so $4$ points: $(0,0)$, $(2,3)$, $(4,6)$, $(6,9)$.

How AMC 10 tests it

• Early: the area of a polygon from its corners, where two lines cross, or the perimeter of a polygon with lattice-point corners.
• "A line through $(a,b)$ cuts a rectangle (or two circles) into equal areas." The line passes through the center of symmetry.
• Middle to late: a line tangent to a circle, or a circle through three points, using the distance formula and completing the square.

Standard approaches

1. Sketch, then write every point, line, and circle as coordinates or an equation before computing.
2. Symmetry first: for a centrally symmetric figure (one that looks the same after a half turn, like a rectangle), every line through its center cuts it into equal areas. So try the line through the given point and the center. For a convex figure like a rectangle, those are the only lines that do it.
3. For tangency, set the center-to-line distance equal to $r$. It is faster than solving a quadratic.
4. For areas, use the Shoelace Theorem. For lattice points, use the gcd on segments and count row by row.

Worked example

The line $y = \frac34 x + b$ is tangent to the circle with center $(2, 1)$ and radius $2$. What is the sum of all possible values of $b$?

(A) $-3$ (B) $-1$ (C) $0$ (D) $1$ (E) $2$

Solution. Step 1. Put the line in the form $ax+by+c=0$. Multiply by $4$: $4y = 3x + 4b$, so $3x - 4y + 4b = 0$.

Step 2. Tangency means the center $(2,1)$ is exactly $2$ from the line:
$$\frac{|3\cdot 2 - 4\cdot 1 + 4b|}{\sqrt{3^2+4^2}} = 2, \quad\text{so}\quad \frac{|2+4b|}{5} = 2 \quad\text{and}\quad |2 + 4b| = 10.$$

Step 3. Either $2+4b = 10$, so $b = 2$, or $2+4b = -10$, so $b = -3$. So there are two tangent lines.

Step 4. The sum is $2 + (-3) = -1$.

The answer is $\boxed{\textbf{(B)}\ -1}$.

Pitfalls

• Sign mistakes when rewriting $y = mx + b$ as $ax + by + c = 0$. Example: $y = 2x+1$ becomes $2x - y + 1 = 0$, not $2x + y + 1 = 0$.
• Forgetting that $|\text{something}| = k$ has two solutions, $k$ and $-k$. That is why a tangent problem usually has two lines.
• Reflecting over $y = x$ by changing signs instead of swapping. $(2,5)$ goes to $(5,2)$, not $(-2,-5)$.

Traps that recur

• Counting lattice points as the area n^2 instead of (n+1)^2, or double counting the shared y-axis column when finding the points of R union S. (2022 AMC 10A #25)
• Taking an interval symmetric around 2/3, or forgetting that points on the line count, so 2/3 itself is the lower endpoint and 19/28 the upper. (2021 AMC 10B #25)
• Using 20 as the radius because the problem says diameter 20, which replaces 10 by 20 in the distance equation and produces g = 80/3, a value that cannot be written in the required form. (2025 AMC 10A #20)
• Summing the three segment lengths only once (8), or treating the region as a parallelogram spanned by just two of the segments. (2023 AMC 10B #24)