rearranging equations into y=mx+c

**vanpopta** · September 17th 2008, 11:50 AM

Hello, i am new to this site, an i was not sure where to post (as i am not in the US schooling system), i am doing AS level maths in the UK if this helps?

i need to rearrange equations to y=mx+c

5x+y=4

x-2y+6=0

x+2y=3

y-6=2(x+4) <-- No idea whatsoever.

I hope you can help me :S

**courteous** · September 17th 2008, 12:01 PM

$5x+y=4 \Rightarrow y=-5x+4$
$x-2y+6=0 \Rightarrow y=\frac{1}{2}x+3$
$x+2y=3 \Rightarrow y=-\frac{1}{2}x+\frac{3}{2}$
$y-6=2(x+4) \Rightarrow y=2x+14$

**11rdc11** · September 17th 2008, 12:02 PM

Here is an example

$x-2y+6=0<br />$

$-2y=-x-6$

$y= \frac{1}{2}x +3$

**11rdc11** · September 17th 2008, 12:03 PM

Originally Posted by courteous

$5x+y=4 \Rightarrow y=-5x+4$
$x-2y+6=0 \Rightarrow y=\frac{1}{2}x+6$
$x+2y=3 \Rightarrow y=-\frac{1}{2}x+\frac{3}{2}$
$y-6=2(x+4) \Rightarrow y=2x+14$

Check your 2nd example

**vanpopta** · September 17th 2008, 12:28 PM

Thank-you so much

Now i shall struggle along with the other questions!!

**vanpopta** · September 18th 2008, 01:18 AM

I got to a later question, which was State whether the following were parallel, perpendicular, or neither, I managed to do the parallel ones, so i need now to know whether these were perpendicular or not

Is $y=3x+4$

perpencicular to

$y=2-\frac{1}{3}x$

and

$x+3y-2=0$

**courteous** · September 18th 2008, 03:23 AM

Originally Posted by vanpopta

I got to a later question, which was State whether the following were parallel, perpendicular, or neither, I managed to do the parallel ones, so i need now to know whether these were perpendicular or not

Is $y=3x+4$

perpencicular to

$y=2-\frac{1}{3}x$

and

$x+3y-2=0$

General equation is $y=kx+n$ , where $k$ is the slope of the line and $n$ is the point of line intersecting the y-axis.
So, the slopes are $y=3x+4 \Rightarrow 3$ ,
$y=2-\frac{1}{3}x \Rightarrow -\frac{1}{3}$ ,
$x+3y-2=0$ (now you know how to rearrange this

) $\Rightarrow -\frac{1}{3}$ .

When $k$ equals the lines are parrarel (or the same), and when $k_1*k_2=-1$ (where the $tan$ is undefined, that is, denominator is 0; more at here) the lines are perpendicular. Now, you only need to compare those ks.

**sqleung** · September 18th 2008, 03:25 AM

For $y = 3x + 4$ , you will notice it has a slope of 3.

Now, a line is perpendicular to the above line if its slope is the inverse reciprocal - ie. slope of $\frac{-1}{3}$

$y = 2 - \frac{-1}{3}x$ does indeed have a slope of $\frac{-1}{3}$ and so, it is perpendicular to the line.

Also:

$x + 3y - 2 = 0$

$x + 3y = 2$

$3y = 2 - x$

$y = \frac{2}{3} - \frac{1}{3}x$

And its slope is also $\frac{-1}{3}$ and hence, it's perpendicular to the line.

**vanpopta** · September 18th 2008, 01:08 PM

Thanks a lot guys

That helped me, for the next 5 questions!

However i got stuck again.

Line q has the equation $y=6x+1$

(a) Give the equation of the line through $(1,0)$ parallel to q
(b) Give he equation of the line through $(3,1)$ perpendicular to q

I think (a) may well be $y=6x-6$

However, i am not confident of that one bit...

**11rdc11** · September 18th 2008, 01:32 PM

Originally Posted by vanpopta

Thanks a lot guys

That helped me, for the next 5 questions!

However i got stuck again.

Line q has the equation $y=6x+1$

(a) Give the equation of the line through $(1,0)$ parallel to q
(b) Give he equation of the line through $(3,1)$ perpendicular to q

I think (a) may well be $y=6x-6$

However, i am not confident of that one bit...

You are correct on a)

Remember

$y_2 - y_1 = m(x_2 - x_1)$

so for a)

$y-0 = 6(x-1)$

$y = 6x + 6$

to do part b) remember that the perpendicular line is the negative reciprocal

$y-1= -\frac{1}{6}(x-3)$

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