# Find the coordinates of the foot of perpendicular from the point (-1,3) to the line $3x-4y-16=0$

Toolbox:
• Slope of a line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\large\frac{(y_2-y_1)}{(x_2-x_1)}$
• If two lines are perpendicular then the product of their slopes is -1.
Let (a,b) be the coordinates of the foot of the perpendicular from the point (-1,3) to the line $3x-4y-16=0$
Slope of the line joining (-1,3) and (a,b) is $m_1 = \large\frac{b-3}{a+1}$
Slope of the line $3x-4y-16$ is
$m_2 = \large\frac{3}{4}$
Since the lines are perpendicular $m_1m_2=-1$
(i.e.,) $\large\frac{b-3}{a+1} $$\times \large\frac{3}{4}$$=1$
$3(b-3)=-4(a+1)$
$\Rightarrow 4a+3b=5$------------(1)
Since the point (a,b) lies on the line $3x-4y=16$
$\qquad 3a-4b=16$------------(2)
On solving eq(1) and (2) we get,
$\qquad (\times 3) 4a+3b=5$
$\qquad ( \times 4) 3a-4b=16$
$\qquad \qquad 12a+9b=15$
$\qquad \quad -12a-16b=-64$
$\qquad \qquad \qquad 25b=-49$
$\Rightarrow b = -\large\frac{49}{25}$
Substituting for b in eqn (1)we get,
$4a+3 \bigg( \large\frac{-49}{25} \bigg)$$=5 \Rightarrow 4a=5+\large\frac{147}{25} \Rightarrow 4a=\large\frac{125+147}{25}$$= \large\frac{272}{25}$
$\therefore a = \large\frac{272}{100}$
$\Rightarrow a = \large\frac{68}{25}$
Hence $a = \large\frac{68}{25}$ and $b=-\large\frac{49}{25}$
Hence the coordinate of the foot of the perpendicular are $\bigg( \large\frac{68}{25}$$, \large\frac{-49}{25} \bigg)$
edited May 24, 2014