Question 186131
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The {{{Slope=m[1]}}} of a line {{{-3x-5y=2}}} is the negative reciprocal--->{{{m[2]=-1/m[1]}}} of the {{{Slope=m[2]}}} of the line that passes thru (2,-6). (*perpendicular lines remember?)
Given line Eqn via Slope-Intercept Form, {{{y=mx+b}}}
{{{5y=-3x-2}}}---->{{{cross(5)y/cross(5)=(-3x-2)/5}}}
{{{y=highlight((-3/5))x-(2/5)}}}--> {{{Slope=m[1]=-3/5}}}
Then,
{{{m[2]=-1/(-3/5)=-1*(5/-3)=5/3}}}
So therfore thru point (2,-6): (Slope-Intercept Form)
{{{-6=(5/3)2+b}}}---->{{{b=-6-10/3=(-18-10)/3=-28/3}}}, Y-intercept
It follows---->{{{highlight(y=(5/3)x-28/3))}}}, ANSWER
Let Fy=0---->{{{0=(5/3)x-28/3}}}---->{{{(5/3)x=28/3}}}
{{{cross(5/3)x/cross(5/3)=(28/3)/(5/3)=(28/cross(3))(cross(3)/5)}}}
{{{x=28/5}}}
Let Fx=0---->{{{y=(5/3)(0)-28/3}}}}---->{{{y=-28/3}}}
Let's plot the points:
{{{drawing(400,400,-4,10,-10,5,grid(1),graph(400,400,-4,10,-10,5,(5/3)x-(28/3)),circle(28/5,0,.10),circle(0,-28/3,.10),blue(circle(2,-6,.16)))}}}--->Line Eqn {{{y=(5/3)x-28/3)}}}----> Perpendicular to {{{y=(-3/5)x-(2/5)}}}->GREEN Line:{{{drawing(400,400,-4,10,-10,5,grid(1),graph(400,400,-4,10,-10,5,(5/3)x-(28/3),(-3/5)x-(2/5)),circle(28/5,0,.10),circle(0,-28/3,.10),blue(circle(2,-6,.16))))}}}
Thank you,
Jojo</pre>