How would i find an equation of a line passing through point B(-2,4) and parallel to the line 3x-2y=6?

Question

I'm stuck can someone help me please?

Answer 1

Hi Adriana.

First lets rewrite the equation in terms of y
3x - 2y = 6
-2y = 6 - 3x
y = (-3x + 6)/2

We know our parallel line will be the same formula - it will just be higher or lower on the graph.
We also know the new line goes through the point (-2,4).

What is y of the first graph where x=-2?
y = (-3(-2) + 6) / 2
y = (6+6)/2 = 12/2
y = 6

So we know that the first line has a point at (-2,6)
Our second line has a point at (-2,4)

This tells us our line is 2 point lower on the graph in the y direction.
therefore:
y2 = y1 - 2

So our second line has the equation:
y = (-3x + 6)/2 + 2

Let's simplify:
y = (-3x + 6)/2 + 4/2
y = (-3x + 6 + 4)/2
y = (-3x + 20)(1/2)
y = -3/2x + 10

Answer 2

First re-write the given line in y=mx+b form you should get,

 

y=3/2x-3

*you do this because the slope of this line should be the same as the slope in your new line in order for it to be parallel.*

Now write a partially finished equation using that slope.  You should get

 

y=3/2x+b

 

 

Now you plug in the values that were given to you x=-2 and y=4 into the partial equation to solve for b (this is your y-intercept) to get your final equation.

4= (3/2)*-2+b

b=7

 

Now re-write your new equation using the y-intercept you just found

 

y=3/2x+7

 

This is your new equation of a line parallel to 3x-2y=6 and passes through the point (-2,4)

 

Hope this helps.

Janet