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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.

3 Answers 3

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

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    @jasper:... :-)2011-06-23
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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.

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    @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.2011-04-19
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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}$