Distances in the Plane

Distance Between Two Points

Take two points $A(x_1, y_1)$ and $B(x_2, y_2)$. Draw a horizontal leg and a vertical leg to make a right triangle. The legs have lengths $|x_2 - x_1|$ and $|y_2 - y_1|$, and the segment $AB$ is the hypotenuse.

When a problem asks for an exact distance, leave a simplified radical. Only use a decimal if the problem asks for an approximation.

Midpoint and Distance to the Axes

The midpoint of a segment is the point halfway between the endpoints. Its coordinates are the averages of the endpoints' coordinates.

Distance from a Point to a Line

The distance from a point to a line is the length of the perpendicular segment from the point to the line. Any other path to the line is longer.

If the answer has a radical in the denominator, rationalize it: $\dfrac{7}{\sqrt2} = \dfrac{7\sqrt2}{2}$.

The Closest Point on a Line

Sometimes you need not just the distance but where the closest point is. That point is the foot of the perpendicular: the place where the perpendicular from $P$ meets the line.

Triangle $PFQ$ has a right angle at $F$, so $PQ$ is its hypotenuse and is longer than the leg $PF$. That is why the perpendicular gives the minimum distance.

Distance Between Parallel Lines

Parallel lines stay the same distance apart everywhere. So pick any one point on one line and find its distance to the other.

If a problem gives you a point on the second line (for example, a ship's position) instead of its equation, you don't need the equation. Just find the distance from that point to the first line.

Lines at a Given Distance

To find lines parallel to $r: Ax+By+C=0$ at distance $d$ from it, keep $A$ and $B$ and look for the new constant $C'$. The shortcut formula gives $\dfrac{|C' - C|}{\sqrt{A^2+B^2}} = d$, so

There are always two such lines, one on each side of $r$.

If the second line is given with different coefficients (like $4x - 2y + 5 = 0$ next to $2x - y + c = 0$), divide to match them before using the shortcut.

Points at a Given Distance and Unknown Parameters

Two common unknowns show up in distance problems.

A point on a line at a given distance from another point. Write the point as $(t, mt + b)$ using the line's equation. Plug it into the distance formula, square both sides, and solve the resulting quadratic.

An unknown coefficient in the line. Put the parameter into the point-to-line formula, set it equal to the given distance, and square both sides. Squaring is safe here because both sides are nonnegative.

Area of a Triangle

Take one side as the base. Its length comes from the distance formula. The height is the distance from the opposite vertex to the line containing that side. Then $\text{Area} = \tfrac12 \cdot \text{base} \cdot \text{height}$.

To find the line through two points, compute the slope $m = \dfrac{y_2-y_1}{x_2-x_1}$, write $y - y_1 = m(x - x_1)$, and rearrange into $Ax+By+C=0$.

Equidistant Points and Loci

A locus is the set of all points that satisfy a condition. The points equidistant from two points $P$ and $Q$ form the perpendicular bisector of $PQ$.

To find a locus, call a general point $(x, y)$, write the distance condition, and square both sides to get rid of the roots.

If you keep $y$ instead of setting it to $0$, the same steps give $3x + y = 9$. That is the equation of the whole perpendicular bisector, and it passes through the midpoint $(2, 3)$.

For a locus defined with a line, use $d(P, r) = \dfrac{|Ax + By + C|}{\sqrt{A^2+B^2}}$ and square both sides. A factor like "twice the distance" gets squared too, so it becomes a factor of $4$.

Equidistant from Two Lines: Angle Bisectors

The points equally far from two intersecting lines lie on the angle bisectors of the angles they form. Two crossing lines make two bisectors.

If a problem adds a second condition, intersect the bisectors with it. For example, with a circle $x^2 + y^2 = 13$ and the bisector $y = 1 - x$: $x^2 + (1-x)^2 = 13$ gives $2x^2 - 2x - 12 = 0$, so $x = 3$ or $x = -2$. The points are $(3, -2)$ and $(-2, 3)$.

Squares and Rhombuses

In a rhombus (and a square) the two diagonals are perpendicular and bisect each other at the center $M$. The area is $\tfrac12 d_1 d_2$, where $d_1$ and $d_2$ are the diagonal lengths. For a square, $d_1 = d_2$.

Start with a quick test: plug the given vertex $A$ into the equation of the given line. There are two cases.

Case 1: $A$ is not on the line $r$ that holds the other diagonal. Use this plan:

1. The diagonal through $A$ is perpendicular to $r$. Its center $M$ is the foot of the perpendicular from $A$ to $r$.

2. The opposite vertex is $C = 2M - A$, and $d_1 = AC = 2\,d(A, r)$.

3. Get $d_2$ from the area, $d_2 = \dfrac{2 \cdot \text{Area}}{d_1}$. For a square, $d_2 = d_1$.

4. The other two vertices are on $r$, at distance $\tfrac{d_2}{2}$ from $M$ on each side.

Case 2: $A$ is on the line. Then the plan breaks down, and the reason is geometric. The diagonal through $A$ meets $r$ only at the center $M$. If $A$ is on $r$, then $A = M$, but a vertex is never the center. So the line cannot hold the diagonal that avoids $A$. For a rhombus whose second diagonal is on $r$, this means no rhombus exists. For a square it usually means the line holds the diagonal that does contain $A$, and then $A$ is one endpoint of that diagonal. The other endpoint can slide along the line, so you describe the answer with a parameter (see the second example below).

Back to the start: you are at $(6, 4)$ and the highway is $3x + 4y - 12 = 0$. Then $d = \dfrac{|3(6) + 4(4) - 12|}{\sqrt{3^2 + 4^2}} = \dfrac{22}{5} = 4.4$ miles.

Summary