Question 1134499


the equation of the line that satisfies the given conditions:

  Contains the point ({{{-6}}},{{{4}}}) and is {{{parallel}}} to the line{{{ x-3y=6}}}

first, recall that {{{parallel}}} lines have {{{same}}} slope

so, find a slope of the line{{{ x-3y=6}}}

{{{ x-3y=6}}}.......solve for {{{y}}}

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

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

{{{y=(1/3)x-2}}}=> slope is {{{m=1/3}}}

since we have point and slope, use point slope formula to find equation


{{{y-y[1]=m(x-x[1])}}}........plug in slope {{{1/3}}} and {{{x=-6}}} and {{{y=4}}} coordinates of given point


{{{y-4=(1/3)(x-(-6))}}}

{{{y-4=(1/3)(x+6)}}}

{{{y-4=x/3+6/3}}}

{{{y-4=x/3+2}}}

{{{y-4-x/3=2}}}

{{{y-x/3=2+4}}}

{{{-x/3+y=62+4}}}=> the final equation in standard form 



{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-6,4,.12),locate(-6,4,p(-6,4)),
 graph( 600, 600, -10, 10, -10, 10, x/3+6, (1/3)x-2)) }}}