Question

find the foot of the perpendicular drawn from the point (-2,3) on the line 12x-5y+13=0.

Answer 1

$\text{The foot of perpendicular from the point (x_1, y_1) to}$

$\text{the line ax+by+c=0 is given by }$

$\boxed{\bf\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-(ax_1+by_1+c)}{a^2+b^2}}$

$\textbf{Given line is 12x-5y+13=0}$

$(x_1,y_1)=(-2,3)$

$\text{Now,}$

$\frac{x+2}{12}=\frac{y-3}{-5}=\frac{-(12(-2)-5(3)+13)}{(-2)^2+3^2}$

$\frac{x+2}{12}=\frac{y-3}{-5}=\frac{-(-24-15+13)}{4+9}$

$\frac{x+2}{12}=\frac{y-3}{-5}=\frac{26}{13}$

$\frac{x+2}{12}=\frac{y-3}{-5}=2$

$\frac{x+2}{12}=2$

$x+2=24$

$\implies\bf\,x=22$

$\text{and}$

$\frac{y-3}{-5}=2$

$y-3=-10$

$\implies\bf\,y=-7$

$\therefore\textbf{The foot of the perpendicular is (22,-7)}$

Find more:

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