# Foot of perpendicular to line

If $M(x_2,y_2)$ is the foot of a perpendicular drawn from $P(x_1,y_1)$ on the line $ax+by+c=0$, then $$\frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=\frac{-(ax_1+by_1+c)}{a^2+b^2}.$$

This is given as a formula in my module without any explanation. I can understand the first equality since the product of the slopes of two perpendicular lines is $-1$. But I cannot understand what $\large\frac{-(ax_1+by_1+c)}{a^2+b^2}$ means and how the last equality holds. Please explain.

$$\frac{x_2-x_1}{a}=\frac{a(x_2-x_1)}{a^2}=\frac{y_2-y_1}{b}=\frac{b(y_2-y_1)}{b^2}$$

implies

$$\frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=\frac{a(x_2-x_1)+b(y_2-y_1)}{a^2+b^2}=\frac{ax_2+by_2-ax_1-by_1}{a^2+b^2}$$

and using the fact that $M$ lies on the original line we have the result.

This works because $\frac{A}{B}=\frac{C}{D}$ implies that each equals $\frac{A+C}{B+D}$.

• How does that implies this? .. how can we add terms like that? :o – Quixotic Apr 19 '11 at 19:06
• Aha,got it ... right from the elementary proportion chapter!! :) – Quixotic Apr 19 '11 at 19:11
• @Jasper:Although this is not my question and his approach could not be use to solve the problem say finding the foot of the perpendicular or image,but still +1. – Quixotic Apr 19 '11 at 19:17
• @Tretwick: Actually this works. +1. Nice solution. I find this much better than my solution :) – Aryabhata Apr 19 '11 at 19:20
• @Moron:Given the point P and an equation of the line how could we find the foot of the perpendicular using this? – Quixotic Apr 19 '11 at 19:44

This comes for the fact that the distance of point $\displaystyle P(x_1, y_1)$ from the line $\displaystyle ax + by + c = 0$ is given by $$\frac{| ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

If $\displaystyle \frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = k$ say, then we have that, by considering the triangle formed by drawing a horizontal line from M and a vertical line from P (depends on your figure, though).

For instance see this figure: $$(ka)^2 + (kb)^2 = \left(\frac{| ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}\right)^2$$ and thus

$$|k| = \frac{|ax_1 + by_1 + c|}{a^2 + b^2}$$

I am guessing you can now determine the right sign to take.

• I don't understand,what I could is that $\frac{| ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$ is the distance of $PM$ right? but I could not understand how the triangle is formed? and also why/how $(ka)^2 + (kb)^2 = (\frac{| ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}})^2$? – Quixotic Apr 19 '11 at 19:00
• @Tretwick: It comes from the Pythagorean theorem. I will try to add a figure.. – Aryabhata Apr 19 '11 at 19:03
• Yes,I guessed that but how are you getting a right angle right-angle triangle,..a figure would be great help :) – Quixotic Apr 19 '11 at 19:04
• Got it.. thanks a lot umm the sign part is due to the fact that slope is perpendicular right? – Quixotic Apr 19 '11 at 19:38
• @Tretwick: Actually I would recommend you go through Jasper's neat answer. I haven't thought about the sign, but for all points (x1,y1) lying on one side of the line ax+by+c = 0, the values ax1 + by1 + c are all of the same sign. You should be able to use that. – Aryabhata Apr 19 '11 at 19:40

Since, the point $M(x_2, y_2)$ i.e. foot of perpendicular lies on the line $ax+by+c=0$ hence, it will satisfy the equation as follows $$ax_2+by_2+c=0$$ $$=> ax_2+by_2=-c$$

Now applying the condition of line $ax+by+c=0$ & the line joining $P(x_1, y_1)$ & $M(x_2, y_2)$ to be perpendicular to each other, we have $$(\text{slope of line: ax+by+c=0})\times(\text{slope of line PM})=-1$$ $$=>\left(\frac{-a}{b}\right)\times \left(\frac{y_2-y_1}{x_2-x_1}\right)=-1$$ $$=>\frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}$$ $$=>\frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=\frac{a(x_2-x_1)}{a^2}=\frac{b(y_2-y_1)}{b^2}=\frac{a(x_2-x_1)+b(y_2-y_1)}{a^2+b^2}$$$$=\frac{-ax_1-by_1+ax_2+by_2}{a^2+b^2}=\frac{-ax_1-by_1+(ax_2+by_2)}{a^2+b^2}=\frac{-ax_1-by_1+(-c)}{a^2+b^2}=\frac{-(ax_1+by_1+c)}{a^2+b^2}$$