Find the point on the line 3 x + y - 5 =0 which is closest to the point ( -3, -3 ).
Easiest way:
If (a,b) is some point off the line
Ax + By + C = 0 , then distance from that point to the line is
|Aa + Bb + C|/√(A^2+B^2)
do distance = |3(-3) + 1(-3)-5|/√(9+1)
= |-9-3-5|/√10
= 17/√10
long way:
slope of line = -3
then slope of perpendicular = 1/3
equation of perp:
y+3 = (1/3)(x+3)
3y + 9 = x+3
x = 3y+6
sub into equation:
3(3y+6) + y - 5=0
9y+18+y-5=0
10y = -13
y = -13/10 = -1.3
x = -39/10 + 6 = 2.1
distance between (2.1, -1.3) and (-3,-3)
= √( 5.1^2 + 1.7^2) = √28.9 = appr.5.3758
answer from above 17/√10 = appr. 5.3758
(All is good!)