Question 174056
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Given Line 1: {{{y=3x+2}}}---->it follows slope-Intercept form {{{y=mx+b}}},
where {{{m=Slope=highlight(3)}}} and {{{b=yIntercept=2}}}.
In Line 1, we designate {{{Slope}}} as {{{m[1]}}}.
when 2 lines are perpendicular, the Slope of the other line is "negative reciprocal",
which means {{{m[2]=-1/m[1]}}}. Therefore, to get the slope of Line 2 that is perpendicular----->{{{m[2]=-1/highlight(3)=-1/3}}}
Then, thru point (1,-1) where {{{x=1}}} & {{{y=-1}}},also we use {{{m[2]=-1/3}}}. Via our Slope-Intercept form:
{{{-1=(-1/3)(1)+b}}}
{{{-1=-1/3+b}}}--->{{{b=-1+(1/3)=(-3+1)/3=-2/3}}}, Y-intercept:
Then the eqn of Line 2 according to our Slope-Intercept Form: {{{highlight(y=(-1/3)x-2/3)}}} or {{{highlight((1/3)(x)+y=-2/3)}}} STandard Form (Answer)
As we sse the graph:
{{{drawing(300,300,-6,6,-6,6,grid(1),graph(300,300,-6,6,-6,6,3x+2,(-1/3)x-2/3))}}}-------> RED, Line 1 : GREEN, Line 2
Thank you,
Jojo</pre>