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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.

Quixotic
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4 Answers4

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

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 $\frac{A+C}{B+D}$.

6

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:

enter image description here

$$(ka)^2 + (kb)^2 = \left(\frac{| ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}\right)^2$$ 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.

Aryabhata
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  • 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
2

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 $$(\text{slope of line: ax+by+c=0})\times(\text{slope of line PM})=-1$$ $$=>\left(\frac{-a}{b}\right)\times \left(\frac{y_2-y_1}{x_2-x_1}\right)=-1$$ $$=>\frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}$$ $$=>\frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=\frac{a(x_2-x_1)}{a^2}=\frac{b(y_2-y_1)}{b^2}=\frac{a(x_2-x_1)+b(y_2-y_1)}{a^2+b^2}$$$$=\frac{-ax_1-by_1+ax_2+by_2}{a^2+b^2}=\frac{-ax_1-by_1+(ax_2+by_2)}{a^2+b^2}=\frac{-ax_1-by_1+(-c)}{a^2+b^2}=\frac{-(ax_1+by_1+c)}{a^2+b^2}$$

Harish Chandra Rajpoot
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Normal of line is given as $\frac{1}{\sqrt{a^2 +b^2}}(a \hat{i} + b \hat{j})$ , distance of point from line is given as $\frac{ax_o + by_o+c}{\sqrt{a^2 +b^2} }$, shift the coordinates such that $P$ is at origin:

$$ X= x-x_p$$ $$Y=y-y_p$$

In this new coordinates, the foot of perpendicular is given as:

$$ <X_{perp},Y_{perp}> = -\frac{(a \hat{i} + b \hat{j})}{a^2 +b^2} (ax_o + by_o +c) $$

Equating components and writing everything in old coordinates:

$$ \frac{x_{perp}- x_p}{a} = \frac{y_{perp} -y_p}{b} = -\frac{1}{a^2 +b^2}(ax_o + by_o +c) $$

tryst with freedom
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