Question

Please please help

Find the value of m if the line 3x+4y-5=0 is parallel to the line (m-2)x-2y+2=0

Answer 1

If the lines are parallel then they have the same slope/gradient value.
One idea is to put each equation into slope-intercept form,
and match up the slopes from there.

Do you understand how to put this into slope-intercept form?\[\large\rm 3x+4y-5=0\qquad\qquad\to\qquad\qquad y=?\]

Answer 2

y=-3x+5/4
right..?

Answer 3

If you mean (-3x+5)/4 then yes.
Brackets are important.
Both terms are being divided by the 4.

Answer 4

So we can clearly see that the slope of the first line is -3/4.
How bout the other one?

Answer 5

not sure-

Answer 6

\[\large\rm (m-2)x-2y+2=0\]We'll subtract 2 to the other side,\[\large\rm (m-2)x-2y=-2\]We'll subtract the x stuff to the other side,\[\large\rm -2y=-(m-2)x-2\]Let's multiply by -1 to get rid of all of the negative stuff,\[\large\rm 2y=(m-2)x+2\]And then divide by 2, ya?

Answer 7

\[\large\rm y=\frac{m-2}{2}x+\frac{2}{2}\]

Answer 8

Confused by any of those steps? :o
What's the slope of this second line?

Answer 9

ahh i get the steps-
would the slope of the second line be
\[\frac{ m-2 }{ 2}\]
or
\[\frac{ -2 }{ 2 }\]

Answer 10

It's the stuff connected to the x, so the (m-2)/2.

Ok great we've found our slopes!
Since these lines are parallel, our slopes are equal,\[\large\rm -\frac34=\frac{m-2}{2}\]From there, solve for m.

Answer 11

m=1/2?

Answer 12

Yayyyy good job \c:/

Answer 13

asdfghjkl thankyou so much !
Oh- is it alright for you to help me with another question?

Answer 14

Sure, I prolly have time for another c:

Answer 15

Find the equation of the line through the intersection of x-y+3=0 and 2x-3y+5=0 with gradient -3/4

Answer 16

x- y+3=0
2x-3y+5=0

So we'd like to know where the lines intersect,
or in other words, the point where they have the same (x,y) value.

We have a couple of options:
We can either solve one of the equations for y and plug it into the other,
this is called `substitution`.

Or we can add/subtract the equations together,
this is called `elimination`.

Are you familiar with either method?

Answer 17

substitution-?

Answer 18

Mmm ok we can try that.
So solve the first equation for y.
y=?

Answer 19

y=x-3?

Answer 20

Ok good.
So the \(\large\rm \color{orangered}{y}\) in our first equation is \(\large\rm \color{orangered}{(x-3)}\).

We want these equations to have the same y value,
\[\large\rm 2x-3\color{orangered}{y}+5=0\]So we're going to let this y in the second equation be the same y from the first, ya?\[\large\rm 2x-3\color{orangered}{(x-3)}+5=0\]We can solve for x from this point.

Answer 21

x=-4?

Answer 22

I'm not sure, I didn't work it out just yet >.<
2x-3x+9+5=0
-x+14=0

Hmm -4 that doesn't look quite right.
I think maybe you forgot to distribute the -3 to the -3, ya?

Answer 23

No actually I think you're right :)
I made a boo boo somewhere.
Lemme see where...

Answer 24

Oh oh, ok.
The first equation should have given us y=(x+3).
That was the issue.

Answer 25

\[\large\rm 2x-3\color{orangered}{(x+3)}+5=0\]

Answer 26

Which yes, will lead to x=-4.

Answer 27

Plug that value into the first equation to find a corresponding y value.

Answer 28

y=-1..?

Answer 29

Ok great!
So we've discovered that these lines intersect at \(\large\rm (x_o,y_o)=(-4,-1)\)

Answer 30

So our new line will pass through this point,
and will have a gradient value of -3/4.

Are you familiar with point-slope form of a line?
\(\large\rm y-y_o=m(x-x_o)\)

Because we have all of the necessary pieces to just plug right in.

Answer 31

it would be y+1=-3/4(x+4) ?

Answer 32

Yayyy good job \c:/