Question

The line that contains the point Q( 1, -2) and is parallel to the line whose equation is

Answer 1

Let's call this line y=mx+C, whereby 'm' will be its gradient and 'C' will be its constant.

If this line is parallel to the line you've just mentioned, it will have a gradient 2/3. We know this, because when we re-arrange the equation you've given us, we get...

$y-4=\frac { 2 }{ 3 } \left( x-3 \right) \\ \\ y-4=\frac { 2 }{ 3 } x-2\\ \\ y=\frac { 2 }{ 3 } x-2+4\\ \\ y=\frac { 2 }{ 3 } x+2$

So, at the moment, our parallel line looks like this...

y=(2/3)*x + C

However, you mentioned that this line passes through the point Q(1, -2). If this is the case, for the line (almost complete) above, when x=1, y=-2. With this information, we can figure out the constant of the line we want to find.

-2=(2/3)*(1) + C

Therefore:

C = - 2 - (2/3)

C = - 6/3 - 2/3

C = - 8/3

This means that the line you are looking for is:

y=(2/3)*x - (8/3)

Let's find out if this is truly the case with a handy graphing app... Well, it turns out that I'm correct.