write eqn of line parallel to y = -1/3x + 5, w/ y-int. of 1

[samthe man]

Write an equation of the line described the line parallel to y = -1/3x + 5, having a y-intercept of 1

 

[jonboy]

Ok lets use slope intercept form:\(\displaystyle \L \;
y = \underbrace m_{slope}x + \underbrace b_{y - {\rm intercept}\)

If a line is parallel it has the same slope.

You know the y - intercept 1.

So plug these into your slope - intercept equation.

 

[jonboy]

Well you can have a negative fraction as it means a downward sloping line. Truse me it's nothing to be perplexed over.

So lets use that slope intercept form:\(\displaystyle \L \;
y = \underbrace m_{slope}x + \underbrace b_{y - {\rm intercept}\)

Since the slope is the same, your new equation has the slope (\(\displaystyle m\)):\(\displaystyle \L \;\,-\,\frac{1}{3}\)

The y - intercept (\(\displaystyle b\)) is:\(\displaystyle \L \;1\)

So subsitute:

\(\displaystyle \L \;
y = \underbrace m_{slope}x + \underbrace b_{y - {\rm intercept}\)

\(\displaystyle \L \;y\,=\,\,-\,\frac{1}{3}x\,+\,1\)

In the future plz don't make duplicate threads. I hope you understand now.

 

[samthe man]

and what happens to the +5

 

[jonboy]

and what happens to the +5

Since the problem says the this new line has a y - interecept of 1, you would not use the previous y - intercept of 5.