Distance of a point from a line

The dis­tance from a point to a line means the short­est dis­tance, which is mea­sured along the per­pen­dic­u­lar from the point to the line. There is a neat for­mula for it, and from it come the dis­tance between two par­al­lel lines, the alti­tude and area of a tri­an­gle, and a test for whether two points lie on the same side of a line. This les­son devel­ops all of these.

The per­pen­dic­u­lar dis­tance for­mula#

Result. The per­pen­dic­u­lar dis­tance from the point $P(x_1, y_1)$ to the line $ax + by + c = 0$ is

$$\displaystyle d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

In words: sub­sti­tute the point into the left side of the equa­tion, take the mod­u­lus, and divide by $\sqrt{a^2 + b^2}$.

Why. Let the line cut the axes at $\displaystyle A\left(-\frac{c}{a}, 0\right)$ and $\displaystyle B\left(0, -\frac{c}{b}\right)$ (assume $a$, $b$, $c$ non-zero; the other cases are sim­pler). The area of tri­an­gle $PAB$ can be found in two ways: as $\displaystyle \frac{1}{2} \times AB \times d$, and from the coor­di­nates by the deter­mi­nant for­mula. Work­ing this out, the area is $\displaystyle \frac{|c|\,|ax_1 + by_1 + c|}{2|ab|}$, while $\displaystyle AB = \frac{|c|\sqrt{a^2 + b^2}}{|ab|}$. Equat­ing the two expres­sions for the area and can­celling gives the for­mula.

Spe­cial case. The dis­tance from the ori­gin to $ax + by + c = 0$ is $\displaystyle \frac{|c|}{\sqrt{a^2 + b^2}}$.

Exam­ple 1#

Find the dis­tance from $(2, 3)$ to the line $3x + 4y - 8 = 0$.

$$\displaystyle d = \frac{|3(2) + 4(3) - 8|}{\sqrt{9 + 16}} = \frac{|6 + 12 - 8|}{5} = \frac{10}{5} = 2$$

Exam­ple 2#

Find $k$ if the dis­tance from $(k, 2)$ to $4x - 3y + 1 = 0$ is $3$.

$$\displaystyle \frac{|4k - 6 + 1|}{5} = 3 \implies |4k - 5| = 15 \implies 4k - 5 = 15 \text{ or } 4k - 5 = -15$$

So $k = 5$ or $\displaystyle k = -\frac{5}{2}$. There are two such points, one on each side of the line.

Dis­tance between par­al­lel lines#

Two par­al­lel lines can be writ­ten with the same $x$ and $y$ coef­fi­cients: $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$. The dis­tance between them is

$$\displaystyle d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$$

(Take any point on one line and find its dis­tance from the other.) Before using the for­mula, make sure the coef­fi­cients of $x$ and $y$ are the same in both equa­tions.

Exam­ple 3#

Find the dis­tance between $3x + 4y - 8 = 0$ and $6x + 8y + 9 = 0$.

Divide the sec­ond equa­tion by $2$ to match the first: $\displaystyle 3x + 4y + \frac{9}{2} = 0$. Then

$$\displaystyle d = \frac{\left|-8 - \frac{9}{2}\right|}{\sqrt{9 + 16}} = \frac{\frac{25}{2}}{5} = \frac{5}{2}$$

Alti­tude and area of a tri­an­gle#

The alti­tude from a ver­tex is the dis­tance from that ver­tex to the line of the oppo­site side.

Exam­ple 4#

In the tri­an­gle with ver­tices $A(1, 3)$, $B(0, 0)$, $C(4, 3)$, find the alti­tude from $A$ and the area.

The line $BC$ passes through the ori­gin with slope $\displaystyle \frac{3}{4}$, so it is $3x - 4y = 0$. The alti­tude from $A$ is

$$\displaystyle \frac{|3(1) - 4(3)|}{\sqrt{9 + 16}} = \frac{|3 - 12|}{5} = \frac{9}{5}$$

$BC = \sqrt{16 + 9} = 5$, so the area is $\displaystyle \frac{1}{2} \times 5 \times \frac{9}{5} = \frac{9}{2}$ square units. (The deter­mi­nant for­mula from the matri­ces course gives the same: $\displaystyle \frac{1}{2}|1(0 - 3) - 3(0 - 4) + 1(0)| = \frac{9}{2}$.)

Which side of a line?#

For the line $ax + by + c = 0$, the expres­sion $ax + by + c$ is zero on the line, pos­i­tive for all points on one side and neg­a­tive for all points on the other side. So:

• two points are on the same side of the line if $ax + by + c$ has the same sign at both;
• they are on oppo­site sides if the signs are dif­fer­ent.

Exam­ple 5#

Are $(1, 1)$ and $(2, -1)$ on the same side of $x + 2y - 2 = 0$?

At $(1, 1)$: $1 + 2 - 2 = 1 > 0$. At $(2, -1)$: $2 - 2 - 2 = -2 < 0$. The signs dif­fer, so the points are on oppo­site sides, and the line passes between them.

Prac­tice#

1. Find the dis­tance from $(3, -2)$ to $4x + 3y + 4 = 0$.
2. Find the dis­tance from the ori­gin to $12x - 5y + 26 = 0$.
3. Find the dis­tance from $(1, 1)$ to $x - y + 2 = 0$.
4. Find the dis­tance between $3x - 4y + 7 = 0$ and $3x - 4y - 13 = 0$.
5. Find the dis­tance between $2x + y - 3 = 0$ and $4x + 2y + 4 = 0$.
6. Find $k$ if the dis­tance from $(1, k)$ to $3x + 4y - 5 = 0$ is $2$.
7. Are $(2, 3)$ and $(-1, 5)$ on the same side of $2x - y + 1 = 0$?
8. Is the ori­gin on the same side of $x + y - 4 = 0$ as the point $(1, 5)$?
9. Find the area of the tri­an­gle with ver­tices $(1, 3)$, $(0, 0)$ and $(4, 3)$ by a dif­fer­ent method from Exam­ple 4: use the alti­tude from $C$ to $AB$.
10. Find the points on the $x$-axis whose dis­tance from $4x + 3y - 12 = 0$ is $3$.

Answers#