<PRE><B>Question 1060097</B>
The perpendicular distance between line 3x-5y=10 and point P (12,6)


Given line has slope  {{{5/3}}}.
Line perpendicular to given line and containing P is of equation  {{{y-6=(5/3)(x-12)}}}.


Where do these two lines meet?
Find intersection of:
-
{{{3x-5y=10}}}
{{{-5y=10-3x}}}
{{{y=-10/5+3x/5}}}
{{{y=-2+3x/5}}}
-
{{{y-6=5x/3-(5/3)12}}}
{{{y-6=5x/3-20}}}
{{{y=5x/3-14}}}
-
Equating expressions for y,
{{{-2+3x/5=5x/3-14}}}
{{{-2+14=5x/3-3x/5}}}
{{{12=5x/3-3x/5}}}
{{{12*15=25x-9x}}}
{{{12*15=16x}}}
{{{x=(12*15)/4}}}
{{{x=45}}}
Use x to evaluate y,
{{{y=5*45/3-14}}}
{{{y=5*15-14}}}
{{{y=61}}}
-
Intersection Point, (45,61).



FIND DISTANCE BETWEEN (45, 61) and (12,6).
{{{highlight_green(sqrt((45-12)^2+(61-6)^2))}}}
{{{highlight(sqrt(4049))}}}



4049 is a prime number.</PRE>