# Dot Product - Distance between Point and a Line

The distance between a point and a line, is the shortest possible distance between the point and the line. The shortest possible distance between a point and a line is the perpendicular distance between the two.

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 have \(\vec{n}\) to be a vector normal to the line, and 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 \(\vec{n}:\)

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

So

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

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 \(Q=(-3,2)\),

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 \((0,0)\).

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:

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

**Cite as:**Dot Product - Distance between Point and a Line.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/dot-product-distance-between-point-and-a-line/