SOLUTION: I understand how to graph when given an equation. I do not understand the reverse. For instance, what is the equation of a line with an intercept of (0,3) and a slope of -3. So

Question 568969: I understand how to graph when given an equation. I do not understand the reverse. For instance, what is the equation of a line with an intercept of (0,3) and a slope of -3. So x is 0 and y is 3. I am assuming the slope would go down to the left along the x axis but I still do not know how to go to the equation. Thanks
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
you are given that a point on the line is (0,3) and the slope is -3.
the slope intercept form of the equation for a straight line is:
y = mx + b
m is the slope and b is the y-intercept.
the point slope form of the equation for a straight line is:
y-y1 = m(x-x1)
x1,y1 is a point on the line.
m is the slope.
you can find the equation using either of these forms.
i will show you how to do it with both.

assume you start with the slope intercept form of the equation.
that form is:
y = mx + b
since the slope you are given is -3, then replace m with -3 to get:
y = -3x + b
you know that a point on the line is (0,3)
replace y with 3 and replace x with 0 to get:
3 = -3*0 + b which becomes:
3 = b
the y-intercept is 3.
replace b in the slope intercept form of the equation with 3 and you get:
y = -3x + 3

assume you start with the point slope form of the equation.
that form is:
y-y1 = m(x-x1)
m is the slope.
x1,y1 is a point on the line.
you know the slope is -3 so you replace m with -3 to get:
y-y1 = -3(x-x1)
you know that a point on the line is (0,3), so you set (x1,y1) = (0,3) and replace y1 in your equation with 3 and x1 in your equation with 0 to get:
y-3 = -3(x-0) which becomes:
y-3 = -3x
add 3 to both sides of this equation to get:
y = -3x + 3

you got the same equation either way as you should.
that's how you do it.

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