Question 62752
The distance from a point (x1,y1) to a line ax + by + c = 0 is given by 

(ax1 + by1 + c)/sqrt(a^2 + b^2)

Here the point is (7,4).

The line is y = -x + 5

==> y + x - 5 = -x + 5 + x - 5 [adding x - 5 to both the sides]

==> x + y - 5 = 0

Here a = 1, b = 1 and c = -5.

Thus the distance between the given point and the given line is 

(7*1 + 4*1 - 5)/sqrt (1^2 + 1^2)

 = (7 + 4 - 5)/sqrt(1 + 1)

= 6/sqrt(2)

multiplying by sqrt 2 in the numerator and the denominator we get,

 = 6*sqrt(2)/sqrt(2)*sqrt(2)

= 6*sqrt(2)/2
= 3 sqrt(2).

Thus the distance = 3sqrt(2) units

Good Luck!!!