# Foot of perpendicular proof.

I know there is few answer on Foot of perpendicular, but my doubt is different, so please don't mark this as the duplicate.

My book says:

Foot of the perpendicular from a point $(x_1,y_1)$ on the line $ax+by+c=0$ is $$\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=-\dfrac{ax_1+by_1+c}{a^2+b^2}$$

My proof:

Slope of line $\perp$ to $ax+by+c=0$ is $\frac{b}{a}\implies\tan\theta=\frac{b}{a}\implies\begin{cases}\cos\theta &=\pm\frac{a}{\sqrt{a^2+b^2}}\\ \sin\theta &=\frac{b}{\sqrt{a^2+b^2}}\end{cases}$

Let $L$ passes through point $(x_ 1,y_1)$ perpendicular to $ax+by+c=0$.

Let $r$ be the algebraic distance of $(x_ 1,y_1)$ from $ax+by+c=0$ $\implies r=\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}$.

Now co-ordinate of any point on $L$ distance $r$ from point $(x_1,y_1)$ can be given as:$$\bigg(x_1+r\cos \theta,\ y_1+r\sin\theta\bigg)$$, where $\theta$ is angle $L$ makes with positive direction of $x$-axis.

Substituting the values:$$\bigg(x_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{a}{\sqrt{a^2+b^2}},\ y_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{b}{\sqrt{a^2+b^2}}\bigg)$$ Now I didn't wrote $\pm$ with $\cos$, whose sign can be calculated as of $\tan$.

Now camparing gives :

$x=x_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{a}{\sqrt{a^2+b^2}}$, $\ \ \ y=y_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{b}{\sqrt{a^2+b^2}}$

Note, I didn't mixed the denominators $\bigg(\sqrt{a^2+b^2}\bigg)$ to $\bigg(a^2+b^2\bigg)$, as I've calculated $\frac{a}{\sqrt{a^2+b^2}}$ from $\tan\theta$ with sign.

When I applied this on some problems gives me the correct answers but is not in the bookish form please help me to do this.

• I read this quickly. Your answer seems to match the book's, so I don't understand what you are asking. – Ethan Bolker Oct 16 '17 at 12:38
• @EthanBolker no, I didn't get on RHS negative sign. – mnulb Oct 16 '17 at 12:40
• OK I see that minus sign. I suspect that the problem is in the "algebraic" distance $r$ you calculate. The sign in the denominator there is $+$ because that's the convention for square roots. Maybe it should be signed appropriately. I'n not posting this as an answer since I haven't checked it. Perhaps someone else will do a careful job. – Ethan Bolker Oct 16 '17 at 12:46
• @EthanBolker try applying it on $(x_1,y_1)\equiv (-3,5)$ on line $x-y+2=0$ on two different forms of my derivation i.e. I-form: $x=x_1+\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{a}{\sqrt{a^2+b^2}}$ and II-form $\dfrac{x-x_1}{a}=\dfrac{ax_1+by_1+c}{a^2+b^2}$. for form I calculate $\cos\theta$ from $\frac{a}{\sqrt{a^2+b^2}}$ with sign (you'll get $-\frac{1}{\sqrt{2}}$). you'll understand, what I want. – mnulb Oct 16 '17 at 12:57

There are many way to find the foot of the perpendicular.The bookish formulae is the same as yours! Just a little bit of reordering; I pick up from where you stopped. $x=x_1-\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{a}{\sqrt{a^2+b^2}}$

$y=y_1-\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\cdot\frac{b}{\sqrt{a^2+b^2}}$

I dont see a reason why you can't mix the denominators.

$x=x_1-\{ax_1+by_1+c\}\cdot\frac{a}{{a^2+b^2}}$

$y=y_1-\{ax_1+by_1+c\}\cdot\frac{b}{{a^2+b^2}}$

$\frac{x-x_1}a=-\frac{ax_1+by_1+c}{a^2+b^2}$

Similiarly; $\frac{y-y_1}b=-\frac{ax_1+by_1+c}{a^2+b^2}$ Or

$$\frac{x-x_1}a=\frac{y-y_1}b=-\frac{ax_1+by_1+c}{a^2+b^2}$$ Note:

The negative sign is due to our assumption that the foot of the perpendicular is below the given point.You probably may have assumed the foot of perpendicular is above the point

Excuse me if i am wrong somewhere;