Question 1155218
The standard form for linear equations in two variables is {{{ax+by=c}}}, and

slope-intercept form is {{{y=mx+b}}}

1. Find an equation of the line containing the point ({{{-2}}},{{{5}}}) with slope {{{m=-3}}};slope-intercept form.

{{{y=mx+b}}}.......plug in a slope and given {{{x}}} and {{{y}}} coordinates of the point

{{{5=-3(-2)+b}}}

{{{5=6+b}}}

{{{b=5-6}}}

{{{b=-1}}}

your equation is:

{{{y=-3x-1}}}



2. Find an equation of the line containing the point ({{{9}}},{{{3}}}) with slope {{{m=-1/3}}}; standard form.

since given slope, start with slope intercept form

{{{3=-(1/3)9+b}}}

{{{3=-3+b}}}

{{{b=3+3}}}

{{{b=6}}}

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

write in standard form:

{{{ax+by=c}}}

{{{(1/3)x+y=6}}} ...both sides multiply by {{{3}}}

{{{ x+3y=18}}}


3. Find an equation of the line containing the ({{{-2}}},{{{-3}}}) with slope {{{m=1/8}}}; slope-intercept form.

{{{-3=(1/8)(-2)+b}}}

{{{-3=-(1/4)+b}}}

{{{(1/4)-3=b}}}

{{{(1/4)-12/4=b}}}

{{{-11/4=b}}}

equation is:

{{{y=(1/8)x-11/4}}}


4. Find an equation of the line containing the point ({{{4}}},{{{0}}}) with slope {{{-3/16}}}; standard form.

since given slope, start with slope intercept form

{{{0=-(3/16)4+b}}}

{{{0=-(3/cross(16)4)cross(4)+b}}}

{{{0=-(3/4)+b}}}

{{{(3/4)=b}}}

standard form of equation is:

{{{y=-(3/16)x+3/4}}}.......both sides multiply by {{{16}}}

{{{16y=-3x+16*3/4}}}

{{{3x+16y=4*3}}}

{{{3x+16y=12}}}