Dot Product - Distance between Point and a Line

The distance between a point and a line, is defined as the shortest distance between a fixed point and any point on the line. It is the length of the line segment that is perpendicular to the line and passes through the point.

The distance $d$ from a point $({ x }_{ 0 },{ y }_{ 0 })$ to the line $ax+by+c=0$ is $$d=\frac { \left\lvert a({ x }_{ 0 })+b({ y }_{ 0 })+c \right\rvert }{ \sqrt { { a }^{ 2 }{ +b }^{ 2 } } } .$$

From the figure above let $d$ be the perpendicular distance from the point $Q({ x }_{ 0 },{ y }_{ 0 })$ to the line $ax+by+c=0.$ We also let $\vec{n}$ be a vector normal to the line that starts from point $P({ x }_{ 1 },{ y }_{ 1 })$.

We can see from the figure above that the distance $d$ is the orthogonal projection of the vector $\vec{PQ}$. Thus we have from trigonometry:

$$d=\left\| \vec { PQ } \right\| \cos\theta .$$

Now, multiply both the numerator and the denominator of the right hand side of the equation by the magnitude of the normal vector $\vec{n}:$

$$d=\frac { \left\| \vec { PQ } \right\| \left\| \vec { n } \right\| \cos\theta }{ \left\| \vec { n } \right\| }.$$

We know from the definition of dot product that $\left\| \vec { PQ } \right\| \left\| \vec { n } \right\| \cos\theta$ just means the dot product of the vector $\vec{PQ}$ and the normal vector $\vec{n}:$

$$\begin{aligned} d&=\frac { \vec { PQ } \cdot \vec { n } }{ \left\| \vec { n } \right\| }\\ \vec{PQ}&=({ x }_{ 0 }-{ x }_{ 1 },{ y }_{ 0 }-{ y }_{ 1 }). \end{aligned}$$

So

$$\begin{aligned} \vec { PQ } \cdot \vec { n } &=({ x }_{ 0 }-{ x }_{ 1 },{ y }_{ 0 }-{ y }_{ 1 })\cdot (a,b)\\ &=a({ x }_{ 0 }-{ x }_{ 1 })+b({ y }_{ 0 }-{ y }_{ 1 }). \end{aligned}$$

And we also have $\left\| \vec { n } \right\| =\sqrt { { a }^{ 2 }+{ b }^{ 2 } } ,$ thus

$$d=\frac { \left| a({ x }_{ 0 }-{ x }_{ 1 })+b({ y }_{ 0 }-{ y }_{ 1 }) \right| }{ \sqrt { { a }^{ 2 }{ +b }^{ 2 } } } =\frac { \left| a({ x }_{ 0 })-a({ x }_{ 1 })+b({ y }_{ 0 })-{ b(y }_{ 1 }) \right| }{ \sqrt { { a }^{ 2 }{ +b }^{ 2 } } }.$$

From the equation of the line we have $c=-a(x_{1})-b(y_{1}),$ which implies

$$d=\frac { \left\lvert a({ x }_{ 0 })+b({ y }_{ 0 })+c \right\rvert }{ \sqrt { { a }^{ 2 }{ +b }^{ 2 } } } .$$

So given a line of the form $ax+by+c$ and a point $(x_{0},y_{0}),$ the perpendicular distance can be found by the above formula.

Find the distance between the line $l=2x+4y-5$ and the point $Q=(-3,2)$,

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From the distance formula we have:

$$d=\frac { \left| 2(-3)+4(2)-5 \right| }{ \sqrt { 2^{ 2 }{ +4 }^{ 2 } } } =\frac { 3 }{ 2\sqrt { 5 } }.$$

Find the distance between the line $l=3x+4y-6=0$ and the point $(0,0)$.

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The distance formula can be reduced to a simpler form if the point is at the origin as:

$$d=\frac { \left| a(0)+b(0)+c \right| }{ \sqrt { a^{ 2 }{ +b }^{ 2 } } } =\frac { \left| c \right| }{ \sqrt { { a }^{ 2 }{ +b }^{ 2 } } } .$$

So we have:

$$d=\frac { \left| -6 \right| }{ \sqrt { 3^{ 2 }{ +4 }^{ 2 } } } =\frac { 6 }{ 5 } .$$