Just general help!

1. Find the equation of a line which passes through the point A(3,-2) and is parallel to line 3y-2x=5

If we start with the first question, the best thing would be to rearrange $3y - 2x = 5$ into the form of $y = mx + c$ so that it's easier to see the gradient.

If a line is parallel to this, what does it mean?

The line will never meet x

Not really, if two lines are parallel, then they have the same gradient this can be seen below

But you are correct that they will never meet each other

So have you rearranged the equation into the form $y = mx + c$ ?

Yeah I got 3y=-2x+5

Can you divide it by 3 so you get the equation for just y

Once you have that, we know that the x coefficient represents the gradient of the straight line.

So, now that we know the gradient of this line, and we know that the gradient of the line parallel to it is the same we can find the equation of the new line by substituting the point A (3,2)

y=-2/3x +5/3

It should be positive $\frac{2} {3}x$


So now for the equation of the parallel line, we currently have $y=\dfrac{2} {3}x + c$
And we know that point A (3,2) lies on this line.

How could we find out the value of c using point A?

Honestly I have no clue x

We know that point A definitely is on this line, so if you substitute in the x and y values at point A into the equation it should work

So $y= \dfrac{2} {3}x + c[br][br]A(3,2) x=3, y=2[br][br]2= \dfrac{2} {3}(3) + c$

Could you tell me what c would be?

Actually, the quickest thing to do is to *keep* the 3y-2x part unchanged. This determines the gradient of the line, changing the constant 'shifts' the line without changing direction. So any parallel line will have the form 3y-2x = C (for some C). Substitute in the point the line has to go through to find the necessary value of C.

Since I know you've been following the "thoughts on the forum" discussion, this is where having a lot more maths experience is a double edged sword. We're a lot quicker to see "actually, the *best* way of doing this is probably...", and at the same time, it's easy for us to suggest things that may be outside a comfort zone. This method would have been a bit of a leap for me when I first sat my A-level, I think.

Nope this makes no sense to me

i'm super sorry, it is quite confusing isn't it

I have actually just been made aware of an easier way to approach this question which should make far more sense.

So if we go back to the start, we have the equation $3y-2x=5$

for the parallel line, we can keep the $3y-2x$ the same as parallel line have the same gradient

then we can substitute the values of point a (3,2) into $3y-2x$

this will find the constant and you will have the equation of the line

does this make slightly more sense? :h