Question 1195347

an equation for a line perpendicular to {{{3y+6x=-6}}} and passing through the point ({{{-4}}},{{{-6}}})

perpendicular lines have a slopes negative reciprocal to each other

so, write the equation of given line in slope intercept form {{{y=mx+b}}}


{{{3y+6x=-6}}}

{{{3y=-6x-6}}}

{{{y=-6x/3-6/3}}}

{{{y=-2x-2}}} => slope is {{{m=-2}}}


then a slope of the perpendicular line will be {{{m=-1/(-2)=1/2}}}

now we have a slope and one point which is enough to find equation of the perpendicular line  using point slope formula


{{{y-y[1]=m(x-x[1])}}}.........plug in {{{m=1/2}}} and point ({{{-4}}},{{{-6}}})


{{{y-(-6)=(1/2)(x-(-4))}}}


{{{y+6=(1/2)(x+4)}}}


{{{y+6=(1/2)x+2}}}


{{{y=(1/2)x+2-6}}}


{{{y=(1/2)x-4}}} -> your line



{{{ drawing( 600, 600, -10, 10, -10, 10, circle(-4,-6,.12),locate(-4,-6,p(-4,-6)),
graph( 600, 600, -10, 10, -10, 10, -2x-2, (1/2)x-4)) }}}