Finding an equation of a line

Find an equation of the line whose y-intercept is 3 and that contains the given point.

1. (3, -3)
2. (4, 5)

1.) x1 = 3
y1 = - 3
x2 = 0
y2 = 3

Slope, m = (y2 - y1) / (x2 - x1)

m = [3 - (- 3)] / (0 - 3)
m = (3 + 3) / (- 3)
m = 6 / - 3
m = - 2

b = y1 - m(x1)

b = - 3 - [- 2(3)]
b = - 3 + 6
b = 3

Equation:

y = - 2x + 3
¯¯¯¯¯¯¯¯¯¯

2.) x1 = 4
y1 = 5
x2 = 0
y2 = 3

m = (3 - 5) / (0 - 4)
m = - 2 / - 4
m = 1/2

b = 3

Equation:

y = 1/2 x + 3
¯¯¯¯¯¯¯¯¯¯¯¯
 

You got it "<3"

1. So the point is (3, -3) and and crosses the y axis at y=3 or (0,3)
Now that you have two points, plug those into the slope formula ((Y2-Y1)/(X2-X1)) to find your slope
(3- (-3))/(0-3)
(3 + 3)/ (-3)
(6/-3)
(-2) = The slope
Now that the slope has been found, plug that into the "point-slope" formula (Y2-Y1 = m(X2-X1)).
Y2-Y1 = -2(X2-X1)
Now for my point, I can choose to plug in either (0,3) or (3, -3). I will choose (0,3) because it's
easier to work with. So
Y2- 3 = -2(X2 - 0)
Y2 - 3 = -2X2 - 0
Y2 = -2x2 + 3 (we can go ahead and now change y2 into just y and x2 into just x.)
y = -2x + 3 (final answer/equation of the line)

2. Alright so this is the same process as 1 except with the point (4,5). Do this one following the above process. For the slope, you should get: 1/2. For the equation you should get: y = (1/2)x + 3


Hope I helped!!!!

y = mx+b

1. -3 = m*3 + 3, so m=-2, and y = -2x + 3

2. 5 = m*4 + 3, so m = 1/2. and y = x/2 + 3

Note: lines 1. and 2. are perpendicular.