Question 189027
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Given line <font color=blue>3x-5y=2</font> in Slope-Intercept Form -->{{{y=mx+b}}};
{{{5y=2-3x}}}, divide by "5" both terms (left & right):
{{{cross(5)y/cross(5)=(2-3x)/5=2/5-(3/5)x}}}
{{{y=highlight(-3/5)x+2/5}}}


Above line Eqn has a {{{Slope=m[1]=-3/5}}}


The Line passing thru point (-1,-3) perpendicular to <font color=blue>3x-5y=2</font> has a {{{Slope=m[2]=-1/m[1]=-1/(-3/5)=(-1)(-5/3)=red(5/3)}}}


Then thru point (-1,-3) via Slope-Intercept Form:
{{{-3=(5/3)(-1)+b}}}
{{{-3=-5/3+b}}} ---> {{{b=-3+5/3=(-9+5)/3=red(-4/3)}}}, Y-Intercept


Let fy=0 ----> {{{0=(5/3)x-4/3}}}
{{{(5/3)x=4/3}}} ----> {{{cross(5/3)x/cross(5/3)=(4/3)/(5/3)=(4/cross(3))(cross(3)/5)}}}
{{{red(x=4/5)}}}, X-Intercept


Therefore, Eqn of the Line is <font color=blue>y=(5/3)x-4/3</font>
In Standard Form and removing fractions:
{{{-(5/3)x+y+4/3=0}}}, multiply whole eqn by 3:
{{{highlight(-5x+3y+4=0)}}}, Answer



We'll see graph:
{{{drawing(400,400,-5,5,-5,5,graph(400,400,-5,5,-5,5,(5/3)x-4/3),circle(4/5,0,.08),circle(0,-4/3,.08),green(locate(1,-.1,x=4/5)),green(locate(0.2,-1.2,y=-4/3)),blue(circle(-1,-3,.08)))}}} --------> perpendicular to{{{drawing(400,400,-5,5,-5,5,graph(400,400,-5,5,-5,5,(5/3)x-4/3,-(3/5)x+2/5),circle(4/5,0,.08),circle(0,-4/3,.08),green(locate(1,-.1,x=4/5)),green(locate(0.2,-1.2,y=-4/3)),blue(circle(-1,-3,.08)))}}}, <font color=green>GREEN, </font><font color=blue>3x-5y=2</font>


Thank you,
Jojo</font>