ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: amanaazim on March 01, 2020, 08:38:57 pm
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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.
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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?
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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
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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?
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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.
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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
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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