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#1 2006-11-02 07:18:57
- SweetKerry
- Member
- Registered: 2006-10-28
- Posts: 3
Help me..
Determine the ratio in which 2y x + 2 =0 divides the line joining (3,-1) and (8, 9).
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#2 2006-11-02 07:23:56
- luca-deltodesco
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- Registered: 2006-05-05
- Posts: 1,470
Re: Help me..
what do you mean by ratio?
you mean the distance along line you gain from incrementing x by 1?
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#3 2006-11-02 08:32:34
- Patrick
- Real Member
- Registered: 2006-02-24
- Posts: 1,005
Re: Help me..
I'd be happy to help you, but I'm not sure either what ratio is supposed to mean in this context
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#4 2006-11-02 21:23:13
- Dross
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- Registered: 2006-08-24
- Posts: 325
Re: Help me..
Determine the ratio in which 2y x + 2 =0 divides the line joining (3,-1) and (8, 9).
I think I know what you mean - take the straight line joining (3, -1) and (8, 9), and cut it where it meets the line 2y - x + 2 = 0, right?
Step 1:
Turn the line 2y - x + 2 = 0, and make y the subject of the equation.
Step 2:
Form an equation in the form y = mx + c for the line that joins the points (3, -1) and (8, 9).
You now have the equations of the two lines in terms of y = <first thing>, and y = <[/i]second thing[/i]>.
Step 3:
See where they meet - set <first thing> = <second thing> and find out what x is. Plug this value of x into the equation of one of the lines (either one, it doesn't matter) and you'll end up with the (x, y) coordinates of the point of intersection.
Step 4:
Use Pythagoras' theorem to find the relevant distances, and hence their ratio.
Is this all clear? Give it a go and do ask if you need more help!
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#5 2006-11-02 23:29:35
- mathsyperson
- Moderator
- Registered: 2005-06-22
- Posts: 4,900
Re: Help me..
You can skip a bit of work by missing out some of Steps 3 and 4.
Once you have the x-value of the intersection point, you can just compare that with the x-values of (3,-1) and (8,9) and get your ratio from that.
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#6 2006-11-03 21:17:58
- som1
- Guest
Re: Help me..
it is quite easy really but if this is homework i shouldn't help
#7 2006-11-03 22:02:05
- MathsIsFun
- Administrator
- Registered: 2005-01-21
- Posts: 7,713
Re: Help me..
it is quite easy really but if this is homework i shouldn't help
Homework hopefully consists of more than one question!
I think it is always best to answer at least one question in a way that helps the person understand.
And I am full of admiration for every person that helps another on this forum.
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