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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
Member
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
Member
Registered: 2006-08-24
Posts: 325

Re: Help me..

SweetKerry wrote:

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: 7,711

Re: Help me..

som1 wrote:

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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