Distance from a point to a line; parallel-line distance

In the previous subtopic we learned how to write a line and read off its slope and intercepts. Now we measure things from a line: the perpendicular distance from a point to the line, and the distance between two parallel lines.

The distance formula

Theorem. The perpendicular distance from a point $P(x_0, y_0)$ to the line $A x + B y + C = 0$ is $\boxed{d = \frac{|A x_0 + B y_0 + C|}{\sqrt{A^2 + B^2}}.}$

Proof sketch. Drop a perpendicular from $P$ to the line, meeting it at $Q$ . The vector $\overrightarrow{PQ}$ is along the normal direction $(A, B)/\sqrt{A^2 + B^2}$ . The signed length of $\overrightarrow{PQ}$ along the normal is $-(A x_0 + B y_0 + C)/\sqrt{A^2 + B^2}$ (computed by substituting $P$ into the linear function and projecting). The unsigned distance is the absolute value.

Distance between parallel lines

If two lines are parallel, write them with the same $A, B$ : $A x + B y + C_1 = 0, \quad A x + B y + C_2 = 0.$ Pick any point on the second line (e.g., $(-C_2/A, 0)$ ) and apply the formula: $d = \frac{|A \cdot (-C_2/A) + 0 + C_1|}{\sqrt{A^2 + B^2}} = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}.$

So $\boxed{d_{\text{parallel}} = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}.}$

A common mistake: if the two equations have different $A, B$ scales (like $2x + 3y = 5$ and $4x + 6y = 11$ ), first scale them to match. The second is $2x + 3y = 11/2$ , so $|C_1 - C_2| = |5 - 11/2| = 1/2$ , $d = (1/2)/\sqrt{13}$ .

Worked examples

Example 1. Distance from $(2, 3)$ to $3 x + 4 y - 10 = 0$ .

$d = \dfrac{|3 \cdot 2 + 4 \cdot 3 - 10|}{\sqrt{9 + 16}} = \dfrac{|8|}{5} = \dfrac{8}{5}$ .

Example 2. Distance from origin to $5 x - 12 y + 26 = 0$ .

$d = \dfrac{|0 + 0 + 26|}{\sqrt{25 + 144}} = \dfrac{26}{13} = 2$ .

Example 3. Distance between $3 x + 4 y - 10 = 0$ and $3 x + 4 y + 5 = 0$ .

$d = \dfrac{|-10 - 5|}{\sqrt{9 + 16}} = \dfrac{15}{5} = 3$ .

Example 4. Find the equation of the line parallel to $4 x - 3 y - 8 = 0$ at distance $2$ from it.

Equations: $4 x - 3 y + c = 0$ . Distance from origin must shift by $2$ : $\dfrac{|c - (-8)|}{\sqrt{16 + 9}} = 2 \Rightarrow |c + 8| = 10 \Rightarrow c = 2$ or $c = -18$ . So two lines: $4 x - 3 y + 2 = 0$ and $4 x - 3 y - 18 = 0$ .

Example 5. Find a point on the line $y = x$ which is equidistant from the lines $4 x + 3 y - 5 = 0$ and $5 x + 12 y - 13 = 0$ .

Let the point be $(t, t)$ . Distances: $d_1 = \dfrac{|4t + 3t - 5|}{5} = \dfrac{|7t - 5|}{5}$ , $d_2 = \dfrac{|5t + 12 t - 13|}{13} = \dfrac{|17 t - 13|}{13}$ .

Equate: $13 |7t - 5| = 5 |17t - 13|$ . Two cases:

(a) $13(7t - 5) = 5(17 t - 13)$ : $91 t - 65 = 85 t - 65 \Rightarrow 6 t = 0 \Rightarrow t = 0$ . Point: $(0, 0)$ .

(b) $13(7t - 5) = -5(17 t - 13)$ : $91 t - 65 = -85 t + 65 \Rightarrow 176 t = 130 \Rightarrow t = 65/88$ . Point: $(65/88, 65/88)$ .

Try it yourself

Distance from $(0, 0)$ to $3 x + 4 y - 25 = 0$ .
Distance from $(1, 2)$ to $x + y - 7 = 0$ .
Distance from $(-3, 5)$ to $5 x - 12 y + 26 = 0$ .
Distance between $2 x - 3 y + 1 = 0$ and $4 x - 6 y - 5 = 0$ .
Distance between $x + y - 4 = 0$ and $x + y + 2 = 0$ .
Find the equation of the line parallel to $3 x + 4 y - 10 = 0$ at distance $5$ .
A line passes through $(2, 3)$ at distance $1$ from origin. Find its slope.
Find the locus of a point equidistant from $x = 3$ and $y = 5$ .
Show that the foot of the perpendicular from origin to $4 x + 3 y - 25 = 0$ is $(4, 3)$ .
Find the distance from $(2, -1)$ to the line through $(1, 2)$ and $(4, -1)$ .
Find the orthogonal distance from $(1, 1)$ to the line $y = x + 4$ .
Find a point on the $x$ -axis equidistant from the lines $y = x$ and $y = -x + 6$ .

Pitfalls / Tricks

Always use the absolute value in the numerator.
For parallel lines, the coefficients of $x$ and $y$ must be identical before subtracting constants.
The denominator $\sqrt{A^2 + B^2}$ never depends on the point.
Insight. The signed expression $\dfrac{A x_0 + B y_0 + C}{\sqrt{A^2 + B^2}}$ tells you not just the distance but the side of the line on which $(x_0, y_0)$ lies , positive on one side, negative on the other.