Distance from a Point to a Line

Distance from a point to a line definition

Let us analyze this problem using the diagram below.

If one was asked to find the distance between the point A and line l then it would be really ambiguous since there are several, in fact, infinite ways we can connect the point and the line.

Hence, one has to be specific about which distance is asked about. Some of the distances are shown by the green segments in the above diagram.

But the only specific distance that can be named and one can quantify is the shortest distance between the point and the line.

The shortest distance between the line and the point is shown by the pink line segment in the above diagram.

The shortest line segment is perpendicular to the line itself. Because any other line segment will make an acute or obtuse angle with the line, the shortest distance will only be possible when it will be perpendicular to the line.

This distance can alternatively be seen as the shortest way by which point A can be brought to line l.

But how can one determine the distance between a point and a line using the equation of a line and the coordinates of the point? Let us explore how we can come up with such a formula.

Distance from a point to a line formula

Let D be a straight line whose equation is given by $ax+by+c=0$ where $a,b$ are not simultaneously 0, and a point A $\left(x_0,y_0\right)$ outside the line, that is not belonging to the line.

The goal is to find the shortest distance between the line D and point P. Let the point where the shortest line segment intersects the line be Q whose coordinates are $\left(x_1,y_1\right)$.

The distance between the point and the line D is the same as the length of the line segment formed by points A and Q or the distance between them. We can use the distance formula to do so but we need to know the coordinates of Q in terms of $x_{0}and{y}_0$ for that purpose.

The distance between a point and a line, StudySmarter Originals

Recall that the gradient of a line with equation $ax+by+c=0$ is given by $m_1=-\frac{a}{b}$. Now the line segment AQ is perpendicular to the line so its slope will be $m_2=\frac{b}{a}$. The reason being that the product of slopes of two perpendicular lines is always -1 that is$m_1{m}_2=-1$.

We now have the slope of the line joining AQ and the coordinates of a point A on it. Using this information, we can now form the equation of line AQ,

$y-y_0=m_2\left(x-x_0\right)$

$y-y_0=\frac{b}{a}\left(x-x_0\right)$

$ay-a{y}_0=bx-b{x}_0$

Since Q lies on this line, we can substitute $\left(x_1,y_1\right)$ for $\left(x,y\right)$ to find the unknowns $x_{1}and{y}_1$.

$a{y}_1-a{y}_0=b{x}_1-b{x}_0$

But Q also lies on the line $ax+by+c=0$$ax+by+c=0$, so it will satisfy the equation of line D, hence we have

$a{x}_1+b{y}_1+c=0$$a{x}_1+b{y}_1+c=0$

The above two lines intersect at Q and hence can be solved simultaneously in order to determine the unknowns $x_{1}and{y}_1$, writing the first equation in terms of $x_1$,

$x_1=\frac{-c-b{y}_1}{a}$

Substituting the expression of $x_1$ in $a{y}_1-a{y}_0=b{x}_1-b{x}_0$, we get

$a{y}_1-a{y}_0=b\left(\frac{-c-b{y}_1}{a}\right)-b{x}_0$$-bc-b^2{y}_1-a^2{y}_1+a^2{y}_0-ab{x}_0=0$

Solving for$y_1$ we get,

$a{y}_1=a{y}_0-\frac{b}{a}\left(c+b{y}_1\right)-b{x}_0$

Expanding the brackets and rearranging the terms, we get

$a{y}_1=\frac{a^2{y}_0-bc-b^2{y}_1-ab{x}_0}{a}$

Multiplying both sides by $a$, we get

$a^2{y}_1+b^2{y}_1=a^2{y}_0-ab{x}_0-bc$

$\left(a^2+b^2\right)y_1=a^2{y}_0-ab{x}_0-bc$

Now we shall divide by $a^2+b^2$, to get

$y_1=\frac{a^2{y}_0-ab{x}_0-bc}{a^2+b^2}$

Substituting this back into $x_1=\frac{-c-b{y}_1}{a}$ to determine $x_1$, we get

$x_1=\frac{-c}{a}-\frac{b}{a}\left(\frac{a^2{y}_0-ab{x}_0-bc}{a^2+b^2}\right)$

Reducing to a common denominator, we get

$x_1=\frac{-c(a^2+b^2)-b{a}^2{y}_0+a{b}^2{x}_0+b^{2}c}{a\left(a^2+b^2\right)}$

Upon simplification, we have

$x_1=\frac{-a^{2}c-b^{2}c-a^{2}b{y}_0+a{b}^2{x}_0+b^{2}c}{a(a^2+b^2)}$

Upon further simplification by eliminating the like terms, we get

$x_1=\frac{b^2{x}_0-ab{y}_0-ac}{a^2+b^2}$

Now we have obtained the coordinates of point Q in terms of the constants we know,

$Q\left(\frac{b^2{x}_0-ab{y}_0-ac}{a^2+b^2},\frac{a^2{y}_0-ab{x}_0-bc}{a^2+b^2}\right)$

Now we can calculate the distance between A and Q using the distance, which is nothing but the distance from the point to the line as we discussed earlier. Let us denote it by d and apply the distance formula,

$d^2={\left(x_1-x_0\right)}^2+{\left(y_1-y_0\right)}^2$

Substituting for $x_1,y_1$ we get

$d^2=\left(\frac{b^2{x}_0-ab{y}_0-ac-a^2{x}_0-b^2{x}_0}{a^2+b^2}\right)+\left(\frac{a^2{y}_0-ab{y}_0-bc-a^2{y}_0-b^2{y}_0}{a^2+b^2}\right)$

Upon further simplification, we get

$d^2=\frac{a^2{\left(a{x}_0+b{y}_0+c\right)}^2+b^2{\left(a{x}_0+b{y}_0+c\right)}^2}{{(a^2+b^2)}^2}$

$d^2=\frac{\left(a^2+b^2\right){\left(a{x}_0+b{y}_0+c\right)}^2}{{\left(a^2+b^2\right)}^2}$

$d^2=\frac{{\left(a{x}_0+b{y}_0+c\right)}^2}{a^2+b^2}$

Taking the square root on both sides, we get,

$d=\pm \frac{a{x}_0+b{y}_0+c}{\sqrt{a^2+b^2}}$

Since d is distance, it cannot be negative so we reject the negative root, giving us,

$d=\frac{a{x}_0+b{y}_0+c}{\sqrt{a^2+b^2}}$

But there is still a possibility when the numerator $a{x}_0+b{y}_0+c$ is negative. To avoid it being negative, its modulus has to be taken,

$d=\frac{\left|a{x}_0+b{y}_0+c\right|}{\sqrt{a^2+b^2}}$

We don’t run into that problem since the denominator is a sum of squares of non-zero numbers, so it will always be positive.

To write the same expression in a more convenient (and easy to remember) form, let us define the equation of the line as $f\left(x,y\right)=ax+by+c$ to get $f\left(x_0,y_0\right)=a{x}_0+b{y}_0+c$, leaving us with,

$d=\frac{\left|f\left(x_0,y_0\right)\right|}{\sqrt{a^2+b^2}}$

Now let us apply this formula through a couple of examples.

Calculating the distance from a point to a line

Calculating the distance from a line to a point is a relatively straightforward process. The equation of a line shall be given and the point to which the distance should be calculated.

Distance from a point to a line example