The line: 3x – 5y + 7 = 0

Comparing ax + by + c = 0 and 3x − 5y + 7 = 0, we get:

a = 3, b = − 5 and c = 7

So, the distance of the point (4, 5) from the straight line 3x − 5y + 7 = 0 is

∴ The required distance is 6/√34