Find the coordinates of the foot of perpendicular from the point $\left(1,3\right)$to the line $3x4y16=0$.

Solution: Given line is $3x-4y-16=0\to \left(1\right)$
We will bring it in general form, $y=mx+c$
$4y=3x-16$
$y=\frac{3}{4}x-4$
So, slope of given line, $m_1=\frac{3}{4}$.
Now, if slope of perpendicular to this line is $m_2$. then,
$m_1\cdot m_2=-1$
$m_2\cdot \frac{3}{4}=-1$
$m_2=-\frac{4}{3}$
Now, equation of this line at point (-1,3),
$\left(y-3\right)=-\frac{4}{3}\left(x+1\right)$
$\Rightarrow 3y-9=-4x-4$
$\Rightarrow 4x+3y-5=0\to \left(2\right)$
Now, we will solve equation (1) and equation (2) to find the point of intersection.
Multiplying (1) with 3, and multiplying (2) with 4 and then adding them,
$9x-12y-48+16x+12y-20=0$
$\Rightarrow 25x=68\Rightarrow x=\frac{68}{25}$
Putting value of $x$ in (1),
$3\left(\frac{68}{25}\right)-4y=16$
$\Rightarrow \frac{204}{25}-4y=16\Rightarrow 100y=-196\Rightarrow y=-\frac{49}{25}$
So, coordinates of foot of perpendicular will be $\left(\frac{68}{25},-\frac{49}{25}\right)$.