Question 686685
In the x-y plane, the equation of a line that is not "vertical" can be written, in the {{{highlight(slope-intercept)}}} form, as {{{highlight(y=mx+b)}}}
The number {{{m}}} is the {{{highlight(slope)}}}, and {{{b}}} is the {{{highlight(y-intercept)}}}.
 
PROBLEM # 1
If a line passes through the point (-1,0) and has the y-intercept {{{b=3}}},
its equation will be {{{y=mx+3}}}
We just have to find {{{m}}}.
THE SIMPLE WAY:
We know that the point (-1,0), with {{{x=-1}}} and {{{y=0}}}
belongs to that line, so we substitute those values for {{{x}}} and {{{y}}} tp get
{{{0=m(-1)+3}}} --> {{{0=-m+3}}} --> {{{highlight(m=3)}}}
The equation for that line is {{{y=mx+b}}} with {{{m=3}}} and {{{b=3}}}, so it is {{{highlight(y=3x+3)}}}
 
PROBLEM # 2
If a line has a slope of {{{m=-2}}},
and passes through the point (3,-7), with {{{x=3}}} and {{{y=-7}}},
the equation of the line in slope-intercept form will be {{{y=-2x+b}}} with some {{{b}}}.
ONE WAY:
We can find {{{b}}} by substituting the coordinates of the point (3,-7) given
{{{-7=-2(3)+b}}} --> {{{-7=-6+b}}} --> {{{-7+6=-6+b+6}}} --> {{{-1=b}}}
So the equation is {{{highlight(y=-2x-1)}}}
ANOTHER WAY:
The equation of a line with slope {{{m}}}.
that passes through point P({{{x[P]}}},{{{y[P]}}}) is
{{{highlight(y-y[P]=m(x-x[P]))}}} in {{{highlight(point-slope)}}} form
Substituting the coordinates of our point P(3,-7),
and {{{m=-2}}}
we get
{{{y-(-7)=-2(x-3)}}} --> {{{y+7=-2x+6}}} --> {{{y=-2x+6-7}}} --> {{{highlight(y=-2x-1)}}}