Question

Find the point on the line 5x+5y+7=0
which is closest to the point (3,−4)

Answer 1

Just use your "closest distance from a given point to a line" formula

D = | 5(3) + 5(-4) + 7 |/√(5^2 + 5^2))
= 2/√50
= 2/5√2
= 2/5√2 units or √2/5 after rationalizing.

or , (the long and traditional way):

you know the slope of the given line is -1
so the slope of a perpendicular from (3,-4) is +1

equation of that perpendicular is:
y+4 = 1(x-3)
y = x - 7
sub into the given line to find the intersection:
5x + 5(x-7) + 7 0
5x + 5x - 35 + 7 = 0
10x = 28
x = 2.8
y = 2.8-7 = -4.2


distance between (3,-4) and (2.8, -4.2)
= √( (3-2.8)^2 + (-4+4.2)^2)
= √(.04+.04)
= √.08 = √(8/100)
= √8/10
= 2√2/10 = √2/5, same as above