Question 1181493


Write the standard form equation of the line that is perpendicular to
{{{6x - 7y - 1 = 0}}} and passes through ({{{0}}}, {{{-2}}}).

first write {{{6x - 7y - 1 = 0}}} in slope intercept form

{{{6x-1 =7y }}}

{{{y=(6/7)x-1/7 }}}=> a slope is {{{m=(6/7)}}}

the line that is perpendicular to given line is negative reciprocal to {{{m=(6/7)}}}

{{{m[p]=-1/(6/7)}}}

{{{m[p]=-7/6}}}


use slope point formula to find equation of the line that is perpendicular to given line

{{{y-y[1]=m[p](x-x[1])}}}........substitute a slope, and coordinates of given point ({{{0}}}, {{{-2}}})


{{{y-(-2)=-(7/6)(x-0)}}}


{{{y+2=-(7/6)x}}}


{{{y=-(7/6)x-2}}}-> write it in standard form {{{ax+by=c}}}

{{{y=-(7/6)x-2}}}.......both sides multiply by {{{6}}}

{{{6y=-7x-12}}}

{{{7x+6y=-12}}}



{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(0,-2,.12), locate(0.2,-2,p(0,-2)),
graph( 600, 600, -10, 10, -10, 10,(6/7)x-1/7 , -(7/6)x-2 )) }}}