Question 189774
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Given Line Eqn {{{system(5x+3y=4(red(EQN1)),3x-5y=3(red(EQN2)))}}}


In {{{red(EQN1)}}}, 
Let Fy=0:
{{{5x+3(0)=4}}} ---> {{{cross(5)x/cross(5)=4/5}}}
{{{red(x=4/5)}}}, x-Intercept (4/5,0)


Let Fx=0
{{{5(0)+3y=4}}} ---> {{{cross(3)y/cross(3)=3/4}}}
{{{red(y=4/3)}}}, y-Intercept (0,4/3)



In {{{red(EQN2)}}},
Let Fy=0:
{{{3x-5(0)=3}}}
{{{3x=3}}} ---> {{{cross(3)x/cross(3)=cross(3)1/cross(3)1}}}
{{{red(x=1)}}}, x-Intercept


Let Fx=0:
{{{3(0)-5y=3}}}
{{{-5y=3}}} --->{{{cross(-5)y/cross(-5)=3/-5}}}
{{{red(y=-3/5)}}},y-Intercept



As we see the graph,
{{{drawing(400,400,-7,7,-7,7,graph(400,400,-7,7,-7,7,(4-5x)/3,(3x-3)/5),circle(4/5,0,.10),circle(0,4/3,.10),blue(circle(1,0,.10)),blue(circle(0,-3/5,.10)))}}} ---> {{{red(RED=EQN1)}}} is perpendicular to {{{green(GREEN=EQN2)}}}




To check, we know the Slope of one line is negative reciprocal of the other,{{{m2=-1/m[1]}}}. We'll see:


In {{{red(EQN1)}}}, 5x+3y=4 ---> {{{3y=-5x+4}}} --->{{{cross(3)y/cross(3)=(-5x+4)/3}}}
{{{y=highlight((-5/3))x+(4/3)}}}, {{{m[1]=-5/3}}}


In {{{red(EQN2)}}}, 3x-5y=3 ---> {{{5y=3x-3}}} --->{{{cross(5)y/cross(5)=(3x-3)/5}}}
{{{y=highlight((3/5))x-3/5}}}, {{{m[2]=3/5}}}


Verifying, {{{m[2]=-1/m[1]}}}
{{{3/5=-1/(-5/3)=(-1)(-3/5)}}}
{{{3/5=3/5}}}


*The two lines are <font color=blue>perpendicular</font> (Answer)





Thank you,
Jojo</font>