Parallel line equations.

RickyOswaldIOW · 2005-12-04 00:57:07

I've figured out how to do this just fine but I have one small problem.
I have problems turning x + y + z = 0 into y = mx + c, i.e:

3x + 2y - 6 = 0
3x + 2y = 6
2y = -3x + 6
y = (-3x+6)/2

I think my problem lies here somwhere.

mathsyperson · 2005-12-04 01:24:14

All of that's right, and then you just have to finish it off by multiplying out of brackets.

y = (-3x + 6)/2

y = -1.5x + 3

RickyOswaldIOW · 2005-12-04 01:27:33

I did indeed get to that but the gradient is supposed to be an integer (whole number).

I have to then work out the equation of the line passing through (2, -3), parralel to y = 3/2x + 3 I work this out as y = 3/2x - 6. The book tells me that the answer is 3x + 2y = 0.

Last edited by rickyoswaldiow (2005-12-04 01:32:43)

Ricky · 2005-12-04 01:49:51

The book is wrong.

3x + 2y = 0 becomes 2y = -3x, and so y = -(3/2)x, which of course has a slope of -3/2, not 3/2.

RickyOswaldIOW · 2005-12-04 02:30:56

The book is often wrong, will you be my A level tutor instead?
To clarify:
"Find the equation for the line parallel to 3x + 2y - 6 = 0 passing through (2, -3)."

so firstly we get the equation of the line we already have in the form y = mx + c:
3x + 2y - 6 = 0
3x + 2y = 6
2y = -3x + 6
y = (-3x+6)/2

y = -1.5x + 3

I'm not sure of the standard formula for the next step but I can work it out in my head by visualising it (since we are only using small and whole numbers). With a gradient of 1.5 that means for every 3 we move up we move 2 along, thus passing the point (2, -3) this line would cross the y axis at -6. Hence the equation for the second line;
y= -1.5x - 6
yet the book tells me : 3x + 2y = 0 (with y on it's own: y = -(3/2)x)

RickyOswaldIOW · 2005-12-04 02:33:25

I think I've been ignoring the - on the gradient and am visualising a reflection of the line.

Ricky · 2005-12-04 02:52:46

Ah, so you did mean a slope of -3/2 after all.

"y= -1.5x - 6"

Try plugging in the point (2, -3) into this. That would be: -3 = -1.5 * 2 - 6, so -3 = -9. That is, of course, wrong. The line can't possibly pass through that point.

It's much easier way to do this problem is called point-slope form. That is, it's the equation that occurs naturally when you have a point and a slope, and you want to form a line (exactly what you have ).

The form is: Slope: m, point: (x1, y1), equation: (y - y1) = m(x - x1). So inputing your values, m = -3/2, x1 = 2, and y1 = -3, you get:

(y - (-3) ) = -3/2 * (x - (2) ), which is y = -3/2x + 3 - 3, which is y = -3/2x. This can then become 2y = -3x, and so 3x + 2y = 0.

No visualization required.

Last edited by Ricky (2005-12-04 02:54:05)

RickyOswaldIOW · 2005-12-04 02:55:59

Brilliant, thanks.

Out of curiosity. Is there any reason why they've chosen to write the answer in the form
3x + 2y = 0
instead of their usual
y = ...

Last edited by rickyoswaldiow (2005-12-04 03:01:04)

Math Is Fun Forum

#1 2005-12-04 00:57:07

Parallel line equations.

#2 2005-12-04 01:24:14

Re: Parallel line equations.

#3 2005-12-04 01:27:33

Re: Parallel line equations.

#4 2005-12-04 01:49:51

Re: Parallel line equations.

#5 2005-12-04 02:30:56

Re: Parallel line equations.

#6 2005-12-04 02:33:25

Re: Parallel line equations.

#7 2005-12-04 02:52:46

Re: Parallel line equations.

#8 2005-12-04 02:55:59

Re: Parallel line equations.

Board footer