Given line is `3x-4y -16 = 0->(1)`
We will bring it in general form, `y = mx+c`
`4y = 3x-16`
`y = 3/4x-4`
So, slope of given line, `m_1 = 3/4`.
Now, if slope of perpendicular to this line is `m_2`. then,
`m_1*m_2 = -1`
`m_2*3/4 = -1`
`m_2 = -4/3`
Now, equation of this line at point (-1,3),
`(y-3) = -4/3(x+1)`
`=>3y-9 = -4x-4`
`=>4x+3y -5 = 0->(2)`
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`
`=> 25x = 68=> x = 68/25`
Putting value of `x` in (1),
`3(68/25)-4y = 16`
`=>204/25-4y = 16 =>100y = -196 => y = -49/25`
So, coordinates of foot of perpendicular will be `(68/25,-49/25)`.
