How Far Is a Point From a Line? One Formula Finds the Shortest Gap

Every non-vertical straight line can be written as $y=mx+c$, and every line of any kind as $Ax+By+C=0$. But the most convenient form depends on what you are given — a point and a slope, two points, or the intercepts.

Once a line is in the form $Ax+By+C=0$, one formula gives the shortest distance from any point to it.

This part covers the forms of a line, converting between them, point-to-line distance and the distance between parallel lines.

**A line parallel to the x-axis is $y=b$ and one parallel to the y-axis is $x=a$; through $(x_1,y_1)$ with slope $m$, the line is $y-y_1=m(x-x_1)$; through two points, it is $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$.

Parallel to an axis.** A line parallel to the x-axis and $3$ units below it is $y=-3$.

Worked example 1 — point-slope. Through $(-2,3)$ with slope $-4$:

$$y-3=-4(x+2)4x+y+5=0$$

Worked example 2 — two-point. Through $(1,-1)$ and $(3,5)$:

$$y+1=\frac{5+1}{3-1}(x-1)=3(x-1)3x-y-4=0$$

Check with $(3,5)$: $9-5-4=0$.

An everyday example. A straight metro line drawn on a city grid map through two station positions is written with the two-point form.

The boundary case. **If $x_1=x_2$, the two-point form fails**; the line is simply $x=x_1$.

**The slope-intercept form $y=mx+c$ shows the slope and y-intercept, the intercept form $\frac{x}{a}+\frac{y}{b}=1$ shows where the line cuts both axes, and rearranging $Ax+By+C=0$ gives either.

Worked example 1 — converting.** For $3x+4y-12=0$:

$$y=-\frac{3}{4}x+3m=-\frac{3}{4},c=3$$
$$3x+4y=12\Rightarrow \frac{x}{4}+\frac{y}{3}=1$$

Worked example 2 — given intercepts. A line cuts the axes at $3$ and $-2$:

$$\frac{x}{3}+\frac{y}{-2}=12x-3y=6$$

Worked example 3 — equal intercepts. A line through $(2,3)$ has equal intercepts $a$:

$$\frac{2}{a}+\frac{3}{a}=1a=5x+y=5$$

An everyday example. Suppose a taxi fare is ₹40 plus ₹12 per km. The fare line $y=12x+40$ is in slope-intercept form: slope $12$, intercept $40$.

The substance. A line through the origin has no intercept form, since both intercepts are $0$.

**The perpendicular distance from $(x_1,y_1)$ to the line $Ax+By+C=0$ is $d=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$, the shortest distance from the point to the line.

Worked example 1.** Distance of $(3,-5)$ from $3x-4y-26=0$:

$$d=\frac{|9+20-26|}{\sqrt{9+16}}=\frac{3}{5}$$

Worked example 2. Distance of the origin from $3x+4y=25$, written as $3x+4y-25=0$:

$$d=\frac{|-25|}{5}=5$$

Worked example 3 — find a constant. The point $(2,3)$ is $2$ units from $3x+4y+k=0$.

$$\frac{|6+12+k|}{5}=2|18+k|=10k=-8\text{ or }k=-28$$

Two parallel lines, one on each side of the point, satisfy the condition.

An everyday example. The shortest walk from a house to a long straight highway is along the perpendicular, and its length is given by this formula on a map grid.

The trap. **Write the line in the form $Ax+By+C=0$ first**; using $3x+4y=25$ with $C=25$ gives the wrong sign inside the modulus.

**For parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ with the same $A$ and $B$, the distance between them is $\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.

Worked example 1.**

$$15x+8y-34=0\text{ and }15x+8y+31=0:d=\frac{|-34-31|}{\sqrt{225+64}}=\frac{65}{17}$$

Worked example 2 — match coefficients first. For $3x-4y+7=0$ and $6x-8y+5=0$, divide the second by $2$ to get $3x-4y+2.5=0$:

$$d=\frac{|7-2.5|}{5}=0.9$$

Worked example 3 — an application. Fahrenheit and Celsius temperatures lie on a straight line through $(0,32)$ and $(100,212)$.

$$F-32=\frac{212-32}{100}(C-0)F=\frac{9}{5}C+32$$

At body temperature $C=37$: $F=66.6+32=98.6$.

An everyday example. A straight canal running beside a parallel road keeps the same gap everywhere, and that gap is the distance between two parallel lines.

The substance. **The $x$ and $y$ coefficients must be identical** before subtracting the constants.