Question

any one help please fan and medal Write the equation of the line that is parallel to the line y = -3x + 12 and passes through the point (-1, 6).

Answer 1

if it's parallel it must have the same slope, -3x

Answer 2

ok sorry do you want the answer choices

Answer 3

sure

Answer 4

ok hold on

Answer 5

Write the equation of the line that is parallel to the line y = -3x + 12 and passes through the point (-1, 6).

y = one thirdx + 7
y = -3x + 3
y = one thirdx + 3
y = -3x + 7

Answer 6

u there

Answer 7

@maddiegirl can u help

Answer 8

Are you familiar with the equation of a line, y=mx+b?

Answer 9

yeah is that slope intercept

Answer 10

Right, then you know that 'm' is the 'slope' when the equation is in that form.

Answer 11

yeah

Answer 12

Two lines that are parallel MUST have the same slope. So in this case, '-3' is your slope that both lines must share.

Answer 13

ok

Answer 14

So to find the new line, all you must do is solve the eqeuation \(y-y_1=m(x-x_1)\)

Answer 15

Where \(y_1\) and \(x_1\) are the points on the new line, and 'm' is the slope of your existing line.

Answer 16

ok so for the y put the2nd y first

Answer 17

can u go deeper into that please

Answer 18

Your 'y' will be a variable as it will represent ANY 'y' on the parallel line, and your '\(y_1\)' will be a point you know. The same applies for 'x' and '\(x_1\)'. Think you can write the starting equation?

Answer 19

let see the y is the 6 and the x is -1

Answer 20

So y-6=.......

Answer 21

Remember you're trying to get this to start: \(y-y_1=m(x-x_1)\)

Answer 22

oh y-6=x-(-1)

Answer 23

You forgot your 'm'

Answer 24

oh yeah srry
y-6=m[x-(-1)]

Answer 25

Exactly, and what is your 'm' in this case?

Answer 26

slope

Answer 27

Yes, but what's its value?

Answer 28

is it 12

Answer 29

Look at your original equation, a parallel line shares the same slope.

Answer 30

-3

Answer 31

Excellent. Now put that all together into the equation and solve for 'y'.

Answer 32

so y-6=-3[x-(-1)]

Answer 33

Yes, exactly.

Answer 34

how do I solve it

Answer 35

oh nvr mind thnk you

Answer 36

You're welcome. :)

Answer 37

r u able to help still @eSpeX

Answer 38

You solved this one, not much more to do.

Answer 39

i mean another one

Answer 40

Sure.

Answer 41

thnx ill post it in another one

Answer 42

Okay, I'll keep an eye out for a mention. :)