Question 166991
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(2,3) and (5,y); perpendicular to a line w/ slope=3/4
Via Point-Slope form: {{{m=(y[2]-y[1])/(x[2]-x[1])}}}
But remember, since it's perpendicular, the slope for above points: {{{m[2]=-1/m[1]}}}
{{{m[2]=-1/(3/4)=-4/3}}}
Subst.,
{{{-4/3=(y[2]-3)/(5-2)}}}
{{{-4/3=(y[2]-3)/3}}}, cross multiply
{{{y[2]-3=(-4/3)(3)=cross(-12)4/cross(3)}}}
{{{y[2]=-4+3=-1}}}
{{{highlight(y[2]=-1)}}}, Answer
We'll check the graph but first let's get the y-intercept:
Via Slope-Intercept: {{{y=mx+b}}}
Thru points: (5,-1)
{{{-1=(-4/3)(5)+b}}}
{{{-1=(-20/3)+b}}}
{{{b=(-1)+(20/3)=(-3+20)/3}}} --> {{{b=17/3}}} y intercept
let's see the graph,
{{{drawing(300,300,-8,8,-8,10,grid(1),graph(300,300,-8,8,-8,10,(-4/3)x+(17/3)),circle(5,-1,.2),circle(2,3,.2),circle(0,17/3,.25))}}} --it passes thru  points(2,3)(5,-1), see also y-intercept @ (17/3)
Thank you,
Jojo</pre>