Question 310277
Put both lines in slope-intercept form, {{{y=mx+b}}}
{{{5y=ax+7}}}
{{{y=(a/5)x+7/5}}}
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{{{by+3x=12}}}
{{{by=-3x+12}}}
{{{y=-(3/b)x+12/b}}}
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Perpendicular lines have slopes that are negative reciprocals,
{{{m1*m2=-1}}}
{{{(a/5)(-(3/b))=-1}}}
{{{(3a)/(5b)=1}}}
{{{3a=5b}}}
{{{highlight_green(a=(5/3)b)}}}