Given line equation: 4x – 3y – 5 = 0

3y = 4x – 5

y = (4/3) x – 5

Slope of the line (m₁) = 4/3

Let the slope of the line perpendicular to the given line be m₂

Then, m₁ x m₂ = -1

(4/3) x m₂ = -1

m₂ = -3/4

Now, the equation of the line having slope m₂ and passing through the point (1, -2) will be

y + 2 = (-3/4) (x – 1)

4y + 8 = -3x + 3

3x + 4y + 5 = 0

Next, for finding the co-ordinates of the foot of the perpendicular which is the point of intersection of the lines

4x – 3y – 5 = 0 …. (1) and

3x + 4y + 5 = 0 …. (2)

On multiplying (1) by 4 and (2) by 3, we get

16x – 12y – 20 = 0

9x + 12y + 15 = 0

Adding we get,

25x – 5 = 0

x = 5/25

x = 1/5

Putting the value of x in (1), we have

4(1/5) – 3y – 5 = 0

4/5 – 3y – 5 = 0

3y = 4/5 – 5 = (4 – 25)/5

3y = -21/5

y = -7/5

Thus, the co-ordinates are (1/5, -7/5)