Find the point on the line 5x+3y−4=0 which is closest to the point (1,5) pls

First put the line in slope intercept form
3y = -5x + 4
y = -5/3x + 4/3

The point which is closest to (1, 5) will be the intersection of the line given and a line perpendicular to it which passes through the point given.

Finding a line perpendicular to the one given means a slope which is the negative reciprocal of the one given.
y = 3/5x + b
Now to solve for b, just plug in the point given
5 = 3/5(1) + b
b = 5 - 3/5
b = 22/5
so
y = 3/5x + 22/5
finding the intersection of these two lines will yield the point we are looking for.
3/5x + 22/5 = -5/3x + 4/3
3/5x + 5/3x = 4/3 - 22/5
34/15x = -46/15
34x = -46
x = -46/34 = -23/17
now to solve for y, just plug x into either of the two lines:
y = 3/5(-23/17) + 22/5
y = -69/85 + 22/5
y = -69/85 + 374/85
y = 305/85 = 61/17
so the point is (-23/17, 61/17)

Hope this helps.