Question 1189115
Two lines 
L1 :  {{{2y-3x-6=0}}}
L2 : {{{3y+x-20=0 }}}

intersect at point A

find the coordinates of the intersection point
{{{2y-3x-6=0}}}.......solve for {{{y}}}
{{{2y=3x+6}}}
{{{y=(3/2)x+3}}}.........eq.1

{{{3y+x-20=0 }}}
{{{3y=-x+20 }}}
{{{y=-(1/3)x+20/3 }}}........eq.2

from eq.1 and eq.2 we have

{{{ (3/2)x+3=-(1/3)x+20/3 }}}......solve for {{{x}}}

{{{ (3/2)x+(1/3)x=20/3 -3}}}

{{{ (11x)/6=11/3}}}

{{{ 11x=6(11/3)}}}

{{{ 11x=2(11)}}}

{{{ x=2(11)/11}}}

{{{ x=2}}}

go to
{{{y=(3/2)x+3}}}.........eq.1, substitute {{{x}}}
{{{y=(3/2)2+3}}} 
{{{y=3+3}}} 
{{{y=6}}} 

=>  point A=({{{2}}},{{{6}}})


A third line L3 is perpendicular to L2 at point A. 
perpendicular lines have slopes negative reciprocal to each other

 line L2 in slope intercept form is {{{y=-(1/3)x+20/3 }}}; so, slope is {{{-1/3}}}

negative reciprocal is {{{-1/(-1/3)=3}}}

=> a slope of the line L3 is {{{3}}}

to find the equation of L3, use a slope point formula

{{{y-y[1]=m(x-x[1])}}} .....plug in a slope and coordinates of the point A

{{{y-6=3(x-2)}}}
{{{y-6=3x-6}}}
{{{y=3x-6+6}}}
{{{y=3x }}}->answer


{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(2,6,.12),locate(2,6,A(2,6)),locate(-2,1.3,L1),locate(-2,6.8,L2),locate(5,5,L3),
graph( 600, 600, -10, 10, -10, 10, 3x , -(1/3)x+20/3,(3/2)x+3)) }}}