my problem is:

"write an equation for the line parallel to the line -1/4y = -4 through the point (1, 8)"

can you help me with this problem please? thanks

Results 1 to 8 of 8

- September 25th 2009, 07:06 PM #1

- Joined
- Sep 2009
- Posts
- 10

- September 25th 2009, 07:10 PM #2

- Joined
- May 2009
- Posts
- 471

When you are given a point (p,q) and a slope=m, you can construct the line with slope m through point (x,y) with the following:

y-q=m(x-p)

So you got m=2 and (p,q)=(-3,-5) so just plug it in

Then you have to change it to y=mx+b, so once you plug your stuff in, multiply through on the right and add q to both sides

y=mx-mp+q

Can you make the correct substitutions?

- September 25th 2009, 07:13 PM #3

- Joined
- Sep 2009
- Posts
- 10

- September 25th 2009, 07:18 PM #4

- Joined
- May 2009
- Posts
- 471

Well if we multiply each side by you get y=16

So the slope=0 cuz 16 is a constant,

So you just need the equation y=8 for the line parallel that goes through (1,8)

Let's make this general

So they give us

So we have found that the parallel slope=

Now we use the point slope form of the line to get (point is (q,p))

Basically, always find the slope of the line given, then use the point thats given to write it as above

- September 25th 2009, 07:32 PM #5

- Joined
- Sep 2009
- Posts
- 10

- September 25th 2009, 07:33 PM #6

- September 25th 2009, 07:36 PM #7

- Joined
- Sep 2009
- Posts
- 10

- September 25th 2009, 10:08 PM #8

- Joined
- Sep 2009
- Posts
- 14

there are various ways....the quick way is that in the equation you can see no mention of x and it is y equal to sumthing therefore it means it is a horizontal line with a gradient of 0

therefore the line parallel would also be horizontal. therefore just take the y value of the point it is passing through and make its equal to y ie

the other way is changee the equation to the form y=mx+c where m is the gradient and c is the y intercept

this is to find the gradient of the parallel line.

therefor the value of m in the parallel line eq would be 0 as well

substitute the x and y values of the point it is passing through to find the value of c

therefore you end up with

which is the same as