Question 964026
{{{-3y+6=-2x}}}
{{{-3y=-2x-6}}}
{{{y=(2/3)x+2}}}


The line is above the origin where x=0.


Find the line containing the origin and having slope  {{{-3/2}}}.
{{{y-0=-(3/2)(x-0)}}}
{{{y=-3x/2}}}.




Find the intersection point of {{{y=-3x/2}}} and {{{y=(2/3)x+2}}}.
{{{-3x/2=2x/3+2}}}
{{{-9x=4x+12}}}
{{{-9x-4x=12}}}
{{{-13x=12}}}
{{{x=-12/13}}}
-
{{{y=-3*(-12/13)/2}}}
{{{3*(12/13)/2}}}
{{{y=3*6/13}}}
{{{y=18/13}}}
-
Intersection point, {{{x=-12/13}}}, {{{y=18/13}}};




Use either Pythagorean Theorem or the Distance formula to find length or distance from (0,0) to  (-12/13, 18/13).


{{{sqrt((12/13)^2+(18/13)^2)}}}


{{{sqrt((144/13^2)+(18^2/13^2))}}}


{{{sqrt((144/13^2)+(324/13^2))}}}


{{{sqrt(468/13^2)}}}


{{{(1/13)sqrt(468)}}}


{{{sqrt(117*4)/13}}}


{{{sqrt(13*9*4)/13}}}


{{{3*2*sqrt(13)/13}}}


{{{highlight(6*sqrt(13)/13)}}}