Question

The coordinate of the foot of the perpendicular from $(x_1,y_1)$ to the line ax+by+c=0 are

• A. $\left(\frac{b^2{x}_1-ab{y}_1-ac}{a^2+b^2},\frac{b^2{y}_1-ab{x}_1-bc}{a^2+b^2}\right)$
• B. $\left(\frac{b^2{x}_1+ab{y}_1+ac}{a^2+b^2},\frac{a^2{y}_1+ab{x}_1+bc}{a^2+b^2}\right)$
• C. $\left(\frac{a{x}_1+b{y}_1+ab}{a+b},\frac{a{x}_1-b{y}_1-ab}{a+b}\right)$
• D. None

Answer 1

Answer: (A)

$\left(\frac{b^2{x}_1-ab{y}_1-ac}{a^2+b^2},\frac{b^2{y}_1-ab{x}_1-bc}{a^2+b^2}\right)$

The perpendicular distance from a point
$(x_1,y_1)=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$
Lets say it meets at $(h,k)$
$\frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-a{x}_1+b{y}_1+c}{a^2+b^2}h=\frac{-a{x}_1+b{y}_1+c}{a^2+b^2}+\frac{x_1}{a}=\frac{-a^2{x}_1+ab{y}_1+ac+a^2{x}_1+b^2{x}_1}{a(a^2+b^2)}h=\frac{b^2{x}_1+ab{y}_1-ac}{a(a^2+b^2)}Similarlyh=\frac{-a{x}_1+b{y}_1+c}{a^2+b^2}+\frac{y_1}{b}=\frac{-ab{y}_1+b^2{y}_1+bc+a^2{y}_1+b^2{y}_1}{b(a^2+b^2)}=\frac{b^2{y}_1-ab{x}_1-bc}{a^2+b^2}$
Therefore the co-ordinates are
$(\frac{b^2{x}_1+ab{y}_1-ac}{(a^2+b^2)},\frac{b^2{y}_1-ab{x}_1-bc}{a^2+b^2})$