Question

What is the equation of a line that is parallel to -x+3y=6 and passes through the point (3,5)?

Answer 1

For this case we have that by definition, the equation of the line in the slope-intersection form is given by:

Where:

m: It's the slope

b: It is the cut-off point with the y axis

By definition, if two lines are parallel then their slopes are equal.

We have the following line:

$-x + 3y = 6\\3y = x + 6\\y = \frac {1} {3} x + \frac {6} {3}\\y = \frac {1} {3} x + 2$

Thus, the slope is:$m_ {1} = \frac {1} {3}$

Then $m_ {2} = \frac {1} {3}$

So, the line is of the form:

$y = \frac {1} {3} x + b$

We substitute the point$(x, y) :( 3,5)$and find b:

$5 = \frac {1} {3} (3) + b\\5 = b$

Thus, the equation is:

$y = \frac {1} {3} x + 5$

Answer:

$y = \frac {1} {3} x + 5$