# Distance Between A Point And A Line

Any Hint on proving that the distance between the point $(x_{1},y_{1})$ and the line $Ax + By + C = 0$ is,

$$\text{Distance} = \frac{\left | Ax_{1} + By_{1} + C\right |}{\sqrt{A^2 + B^2} }$$

What do I use to get started? All I know is the distance formula $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$.

Kindly Help

• Try to find the global minima of your distance formula when plugging in the line for $x_2,y_2$ Nov 26, 2011 at 12:43
• I have no idea how to find the global minima.
– alok
Nov 26, 2011 at 12:55
• A geometric way of looking at it: your problem is equivalent to finding the radius of a circle that is tangent to a given line and centered at a given point. Remember that tangency of a circle and a line corresponds to the resulting quadratic equation having a double root. Nov 26, 2011 at 13:49
• @J.M. Assuming the OP knows the tangency condition (or in the worst case to get tangent replace $x^2$ by $xx'$ etc), this is the best way, IMHO.
– Tapu
Nov 26, 2011 at 14:05

Here is an elementary geometric derivation of the formula:

Any (st.)line perpendicular to the line $$Ax+By+C=0\qquad\text{(1)}$$ is given by $$Bx-Ay+C'=0\qquad\text{(2)}$$ Since (2) has to pass through the point $(x_1,y_1)$ (WHY?), we have $C'=Ay_1-Bx_1$. So, (2) becomes $$Bx-Ay+Ay_1-Bx_1=0\Rightarrow \frac{x-x_1}{A}=\frac{y-y_1}{B}=t\text{ (say)}\qquad(3)$$ From (3), $x=At+x_1$ and $y=Bt+y_1$. This is (called) the parametric equation of the line (2). Each $t$ correspond a point in it and vice-verse. Our next task is to determine the value of $t$ such that (1) and (2) meet at that point. To do so, substituting the value of $x$ and $y$ in (1), we get $t=-\frac{Ax_1+By_1+C}{A^2+B^2}$. Hence the required distance is $$\sqrt{(x-x_1)^2+(y-y_1)^2}=\sqrt{A^2t^2+B^2t^2}=|t|\sqrt{A^2+B^2}=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.$$

For what it's worth, here's the outline using the "derivative approach". Since you tagged this as analytic geometry, I would think the splendid answers by Americo and Didier would be the better approach.

Rewriting your equation of the line for $B\ne0$ gives: $$y={-Ax-C\over B}.$$ (If $B=0$, the line is vertical, and the problem is simple.)

If the point $(x, {-Ax-C\over B})$ is on the line, its distance to $(x_1,y_1)$ is $$\tag{1}D=\sqrt{ (x_1-x)^2 + {\Bigl(y_1- { \textstyle{-Ax-C\over B}}\Bigr)^2 }}.$$

You want to find the smallest value of $D$. This is equivalent to finding the smallest value of $$P(x)=D^2= (x_1-x)^2 + {\Bigl(y_1- { \textstyle{-Ax-C\over B}}\Bigr)^2 }$$

Now, it should be obvious that the distance from a point $p$ on the line to $(x_1,y_1)$ is big, if $p$ is "on an extreme end of the line". So, the minimum distance is attained at a point where $P'(x)=0$.

So, you need to find $P'(x)$, then solve $P'(x)=0$. I'll leave this for you...

You should only get one solution to $P'(x)=0$. The minimum of $P$, and hence the minimum of $D$, would then have to occur at this point . If you then plug this solution into (1), your formula will result after a bit of simplification.

• Just doing it to buy time!!I see there are many methods to prove this.The elementary geometric derivation was the easiest for me,However I have considered all the solutions to be very usefull.Thanks to all of you.
– alok
Nov 26, 2011 at 13:46
• I thought I'd give this a go. I found $P'(x)$ and solved $P'(x) = 0$. Plugging that into (1) was where I had to give up though. I started to work through the formula and things got too complicated for me. Perhaps I'm missing a trick on how to simplify this expression by hand effectively. May 9, 2013 at 10:18
• Following up from my last comment, I asked a separate question about doing the algebraic simplification and got a good response. May 12, 2013 at 15:47

The distance $d$ between the point $P_1(x_{1},y_{1})$ and the line $r$ whose equation is $Ax+By+C=0$ can be derived algebraically as follows:

i) Find the equation of the straight line $s$ passing through $P_1$ and which is orthogonal to $r$. Call $P_2$ the intersecting point of $r$ and $s$.

ii) Find the co-ordinates of $P_2(x_{2},y_{2})$.

iii) Find the distance from $P_1$ to $P_2$. This distance is $d$.

• orthogonal means perpendicular right? If that's the case i know that $m_{1}m_{2} = -1$ or $m_{1} = \frac{-1}{m_{2}}$ where m is the slope.Still strugling to get started.
– alok
Nov 26, 2011 at 13:35
• It looks I have done all the exercises you prescribed for the OP :)
– Tapu
Nov 26, 2011 at 14:10
• @alok Yes. Let $A'x+B'y+C'=0$ be the equation of $s$. Compute the slopes $m=-A/B$ and $m'=-A'/B'$. Let $\theta$ be the angle between $r$ and $s$. From the condition $\tan \theta=0$, and since $P_1$ is a point of $s$ one can show that the equation of $s$ is $B(x-x_1)-A(y-y_1)=0$. Nov 26, 2011 at 14:21
• @Swapan: That's right. In my comment to OP I try to indicate how your equation $(2)$ may be derived using trigonometry. Nov 26, 2011 at 14:29

Assume $(x_0,y_0)$ is on the line and $(x_1-x_0,y_1-y_0)$ is orthogonal to the line. Pythagore says the square of the distance from $(x_1,y_1)$ to any $(x,y)$ on the line is the sum of the squares of the distances from $(x_1,y_1)$ to $(x_0,y_0)$ and from $(x_0,y_0)$ to $(x,y)$. Hence this distance is minimal when $(x,y)=(x_0,y_0)$.

What is $(x_0,y_0)$? First, $(x_0,y_0)$ is on the line hence $Ax_0+By_0+C=0$. Second, the vector $(x_1-x_0,y_1-y_0)$ is orthogonal to the line hence it is proportional to $(A,B)$.

Thus $x_0=x_1-Az$ and $y_0=y_1-Bz$ for some $z$. Plugging this into the first condition yields $A(x_1-Az)+B(y_1-Bz)+C=0$, that is, $(A^2+B^2)z=Ax_1+By_1+C$.

Finally, the distance $D$ from $(x_0,y_0)$ to the line is simply the Euclidean norm of the vector $(Az,Bz)$, hence $D^2=(A^2+B^2)z^2$, and I will let you finish from here.

Consider the line $$\overleftrightarrow{l}$$ whose equation is $$Ax+By+C=0$$, and the point $$P(x_1,y_1)$$ as shown above. Suppose $$\overleftrightarrow{l}$$ intersects the $$x$$ and $$y$$ axis at points $$M$$ and $$N$$ respectively.

To find the distance between $$P$$ and $$\overleftrightarrow{l}$$, let’s first find the distance between the $$x$$ and y-intercepts of $$\overleftrightarrow{l}$$.

So, the length $$MN$$ is given as $$\left| \dfrac{C}{AB}\right| \sqrt{A^{2}+B^{2}}$$

We found the length of $$MN$$ as $$\left| \dfrac{C}{AB}\right| \sqrt{A^{2}+B^{2}}$$ units. Now, suppose $$PQ$$ is the perpendicular drawn from $$P$$ to $$\overleftrightarrow{l}$$ as shown above.

We know that area of a triangle is given as $$Area =\dfrac{1}{2}× Base × Height$$

So, using this result, we can write: $$Area \triangle{PMN}=12×MN×PQ$$

$$\Rightarrow Area \triangle{PMN}=\dfrac{1}{2}× \left| \dfrac{C}{AB}\right| \sqrt{A^{2}+B^{2}} ×PQ$$ $$\Rightarrow PQ=2\times\left| \dfrac{AB}{C}\right| \dfrac{1}{\sqrt{A^2+B^2}} \times Area \triangle{PMN}$$

We found the distance $$PQ$$ as $$PQ=\left| \dfrac{2AB}{C\sqrt{A^{2}+B^{2}}}\right| Area △PMN$$

Now Consider the above △PMN again; We know that the area of a triangle having vertices at $$(x_1,y_1),(x_2,y_2)$$ and $$(x_3,y_3)$$ is given as

$$Area = \frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$$

So, using the above result, Area $$\triangle{PMN}$$ becomes $$\dfrac{1}{2}\left| \dfrac{C}{AB}\right| \cdot \left| Ax_1 +By_1 + C\right|$$ sq. units.

We found the following:

• $$P Q=\left|\frac{2 A B}{C \sqrt{A^{2}+B^{2}}}\right|$$ Area $$\triangle P M N \quad \ldots(\mathrm{i})$$
• Area $$\triangle P M N=\frac{1}{2}\left|\frac{C}{A B}\right| \cdot\left|A x_{1}+B y_{1}+C\right| s q .$$ units $$\cdots (2)$$

From (i) and (ii), we'll get: $$P Q=\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}$$

Therefore, the perpendicular $$distance (d)$$ between the point $$P(x_1,y_1)$$ and the line $$Ax+By+C=0$$ is given as $$d=\dfrac{\left | Ax_{1} + By_{1} + C\right |}{\sqrt{A^2 + B^2} }$$

I am not solving but i'm giving steps for it. 1.draw a perpendicular on giving line from given point. 2.since we've line equation so we can find out its slope(m). 3.use the relation to find out slope of perpendicular line m1*m2=-1 (m1 and m2 are slopes of perpendicular lines).now we have slope of perpendicular line and 1 point from which it is passing by(given point(x1,y1)). 4.find out equation of perpend.line. 5.find out intersection point of bot line. 6.finally calculate distance between given point and intersection point.

(I have edited the answer, as Bob Dobbs pointed out, Case $$B=0$$ need to be considered separately.)

Here is a derivation using vectors.

Let $$P$$ be the point $$(x_1, y_1)$$.

The line with equation $$Ax + By + C = 0$$ has normal vector

$$\boldsymbol{N} = \begin{bmatrix} A \\ B \end{bmatrix}.$$

Note that this is just the coefficients of the equation in order.

Let's pick some point on the line. If $$B \neq 0$$, then when $$x=0$$, $$y=-C/B$$. Let this point on the line, $$(0, -C/B)$$ be $$Q$$.

Consider the vector $$\overrightarrow{QP}= \begin{bmatrix} x_1 \\ y_1 + C/B \end{bmatrix}.$$The distance from point $$P$$ to line $$Ax + By + C = 0$$ is the length of the vector projection of $$\overrightarrow{QP}$$ on $$\boldsymbol{N}$$. This vector projection has length $$\left |\frac{\boldsymbol{N}}{||\boldsymbol{N}||} \cdot \overrightarrow{QP}\right | = \left|\frac{Ax_1 + By_1 + C}{\sqrt{A^2+B^2}}\right|.$$

If $$B=0$$, then the equation of the line becomes $$x = -\frac{C}{A}$$. The distance between $$P$$ and the line is $$\left|x_1 -\frac{C}{A}\right|.$$

Note that the formula is still true. If $$B=0$$ then $$\left|\frac{Ax_1 + By_1 + C}{\sqrt{A^2+B^2}}\right| = \left|x_1 -\frac{C}{A}\right|.$$

Nothing to see here, just me preparing for my Calc 3 final by attempting this.

We can Lagrange this. Let $$f(x) = \sqrt{(x - x_1)^2 + (y - y_1)^2}$$ be the distance of $$(x_1, y_1)$$ from some $$(x, y)$$ on the line $$Ax + By + C = 0$$. Also, let $$g(x) = Ax + By + C = 0$$ be our constraint function, and $$(f(x, y))^2 = (x - x_1)^2 + (y - y_1)^2$$ be our target function. We want to minimize the distance of the point to the line, basically.

Observe that $$\nabla (f(x, y))^2 = \langle 2(x - x_1),2(y - y_1) \rangle$$, and $$\nabla g(x, y) = \langle A, B \rangle$$. Because of the way the gradient behaves around critical points, $$\nabla (f(x, y))^2 = \nabla g(x, y)$$, so $$\langle 2(x - x_1),2(y - y_1) \rangle = \lambda \langle A, B \rangle$$. The solutions to this equation are $$x = x_1 + \dfrac{\lambda A}{2}$$ and $$y = y_1 + \dfrac{\lambda B}{2}$$ (this is nice because then $$x - x_1 = \dfrac{\lambda}{2}A$$ and $$y - y_1 = \dfrac{\lambda}{2}B$$, both of which are helpful later). Plug this back into $$Ax + By + C = 0$$, and you obtain $$\lambda = \dfrac{2Ax_1 + 2By_1 + 2C}{A^2 + B^2}$$.

The distance $$f(x) = \sqrt{(x - x_1)^2 + (y - y_1)^2} = \sqrt{\left(\dfrac{\lambda A}{2}\right)^2 + \left(\dfrac{\lambda B}{2}\right)^2} = \dfrac{|\lambda|}{2} \sqrt{A^2 + B^2}$$. Putting in the expression for $$\lambda$$, the closest distance is $$\min f(x) = \dfrac{1}{2}\left|\dfrac{2Ax_1 + 2By_1 + 2C}{A^2 + B^2}\right|\sqrt{A^2 + B^2} = \dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}},$$as desired.

I like s114's idea and answer but what if $$B=0$$?

My try: A normal vector of the given line $$l$$ is $$\vec n=\langle A,B\rangle$$. Let $$X(x,y)$$ be the the foot of the perpendicular from $$P(x_1,y_1)$$ to $$l$$. Then $$\langle x-x_1, y-y_1\rangle=k\langle A,B\rangle$$ for some $$k\in\Bbb R$$. Hence, we have $$x=kA+x_1$$ and $$y=kB+y_1$$. On the other since $$Ax+By+C=0$$ and we get $$k=-\frac{Ax_1+By_1+C}{A^2+B^2}$$. Finally, the distance of $$P$$ to the given line is $$d=\sqrt{(x-x_1)^2+(y-y_1)^2}=|k|\sqrt{A^2+B^2}=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.$$