What is the equation of the line crossing the y-axis at (0,6) and passing through (4,-2)?

Question

what is the equation of the line crossiing the y-axis at (0,6) and passing through (4,-2)?

Answer 1

Writing the equation is in  slope-intercept form, y=mx+b,   the point (0.6) gives us the y=intercept (the value of y when x=0) as b=6.   So, y=mx + 6.
 
Putting in the point (4,-2),    -2 = m*4 + 6   
                                           -8 = m*4             (subtract 6 from both sides)
                                            -2 = m                (divide both sides by 4)

The equation of the line is     y = -2x + 6

Checking (very important):
       Is   6 = -2(0) + 6        yes.
       Is   -2 = -2(4) +6        yes.

Answer 2

Nicole,

 

The question is begging for us to put the equation of the line in slope-intercept form: y=mx+b

 

b is the y-intercept--the value of y where the line crosses the y-axis.  The y-axis is the line x=0.

If we substitute x=0 into our linear equation, we obtain y=m(0)+b=b as expected.

 

In the question we are told the line crosses the y-axis at the point (0,6).  In fact, this is a redundant statement, because the point (0,6) occurs when x=0...on the y-axis.  The y-value is 6 and so the y-intercept is 6.  b=6.

 

m is the slope of the line.  Slope is change in y/change in x...or (y2-y1)/(x2-x1)=(-2-6)/(4-0) obtained from the two points given in the question.  m=-8/4=-2.

 

Therefore the equation of the line we require is y=-2x+6

Answer 3

This problem lends itself to the slope-intercept form of a line, y = mx + b. where m is the slope (rise/run) and b is the y intercept, 6, where the line crosses the y axis.

 

So far we have y = mx + 6.

 

We still need to know the slope, m.

 

m = rise over run:

 

rise = [6 - (-2)] = 10

 

run = 0 - 4 = -4

 

m = 10/(-4) =  - 2.5

 

 

So we can write the equation of the line as y = -2.5x + 6