Question 749603
A line containing (-6,-3) and is perpendicular to y=3x+5 also contains the point on both lines; this is the point closest to (-6,-3) on y=3x+5.


You want the line {{{y=-(1/3)x+b}}} which contains (-6,-3).  
{{{b=y+(1/3)x}}}
b=-3+(1/3)(-6)
b=-3-2
{{{b=-5}}}.
The perpendicular line is {{{y=-(1/3)x-5}}}


What is the intersection of {{{y=-(1/3)x-5}}} with {{{y=3x+5}}} ?

{{{-(1/3)x-5=3x+5}}}
{{{-3x-(1/3)x=10}}}
{{{3x+(1/3)x=-10}}}
{{{9x+1x=-30}}}
{{{10x=-30}}}
{{{x=-3}}}
Substituting this {{{y=3(x)+5}}}
{{{y=3(-3)+5}}}
{{{y=-9+5}}}
{{{y=-4}}}.


That is the point (-3,-4).  This is the point on BOTH lines, shortest distance on the y=3x+5 to (-6,-3).

How far is (-3,-4) from (-6,-3)?


Distance:  {{{sqrt((-3-(-6))^2+(-4-(-3))^2)}}}
{{{sqrt(3^2+1^2)}}}
{{{highlight(sqrt(10))}}}------------this is how far is y=3x+5 to (-6,-3).