parallel line equation help

[amanaazim]

Can someone please explain how to do this please?

Question: The equation of the line that passes through the point (5,9) and is parallel to the line : y=3x+7 is.

[Evolio]

Hey!
So the equation that is parallel means that it has the same gradient as the line. So the equation that you're trying to find has equation: y=3x+c. Then you substitute the point (5,9) where x=5 and y=9 and that should give you the equation. 

What specific part were you struggling with?

[Rose34]

Can someone please explain how to do this please?

Question: The equation of the line that passes through the point (5,9) and is parallel to the line : y=3x+7 is.

You know how the general equation is Y=mx+c
now that it is saying that it is parallel which means that its gradient is the same thus m is the same for the new equation would be 3
now you sub 3 in the general form so it becomes: y=3x+c
since it gave you the point (5,9), you have to sub it back in to find c:  y=3x+c which will be 9=3(5)+c
9=15+c
9-15=c
c=-6
* now you are almost done, you just need to put everything together into the equation of y=mx+c
c=-6
M=3
thus
Y=3x-6

I hope that helps

[amanaazim]

Hey!
So the equation that is parallel means that it has the same gradient as the line. So the equation that you're trying to find has equation: y=3x+c. Then you substitute the point (5,9) where x=5 and y=9 and that should give you the equation. 

What specific part were you struggling with?

hey i'm just new to this so i don't know about how messages and stuff work so hope this gets sent through. i understood what you meant but where did the 7 go?

[colline]

hey i'm just new to this so i don't know about how messages and stuff work so hope this gets sent through. i understood what you meant but where did the 7 go?

To put simply, sometimes questions would give you 'distracting' information which wouldn't actually be needed for you to solve for the answer.

In this case, the 7 of the original equation is one of these 'distracting' pieces of information - you don't actually need it. The only useful info you need from the equation given is the gradient.

[Bri MT]

hey i'm just new to this so i don't know about how messages and stuff work so hope this gets sent through. i understood what you meant but where did the 7 go?

Hey,

You've replied fine

The only information you need from the original line is its gradient (3) - the y-intercept (7) isn't needed.

This is because the gradient alone determines whether two lines are parallel



Edit: beaten by colline

[TheEagle]

hey i'm just new to this so i don't know about how messages and stuff work so hope this gets sent through. i understood what you meant but where did the 7 go?

Welcome to the forums, I reckon posting in the Methods thread is more convenient but all good!

Attached below is a digram of both lines showing that they are parallel despite the c