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fade out · 26-06-2012, 16:37

Can someone help I can't figure out where I have gone wrong here. It is probably staring at me in the face but I have got brain freeze or something!

A(4,-3) and B(-2,0) are the end points of a line segment. The point P(2,-2) divides [AB] internally in the ratio h:k. Find h:k.

I have subbed in the co ordinates into the formula for internal division and then let each coordinate equal to a coordinate in P(2,-2).

The equations I get from this however are no good as I seem to end up with -4h+2k=0 and 2h-k=0.

Can someone point where I have gone wrong?

im invisible · 26-06-2012, 16:43

just take either the x or the y values, Y makes it easier

A -3
P -2
B 0

so thats 1:2

Delphi91 · 26-06-2012, 16:52

You could always use the "length of a line" formula to calculate |AP| and |PB|. Let H = |AP| and K=|PB|. Put them together in a ratio and simplify.

fade out · 26-06-2012, 16:52

Thanks invisible! Is there a method using the formula though?

Also your method works too Delphi thank you!

**MathsManiac** · 27-06-2012, 00:12

You didn't go wrong. Either one of your two equations tells you that k=2h, so that tells you that h:k is 1:2, or 2:4, or 3:6, etc., which all represent the same ratio, as long as the second number is twice the first.

(The fact that your two equations are equivalent just confirms that the given point is indeed on the line AB)

fade out · 28-06-2012, 23:29

Thanks maths maniac makes sense now! I'm onto another problem now!

Find the equation of the line such that the x-axis bisects the angle between it and the line 2x-3y-6=0

mikeystipey · 28-06-2012, 23:55

Try and give us your attempt first OP - here's a starting point for one way to do it: Find 2 points on the line 2x-3y-6=0 by subbing in x=0 and y=0 respectively, giving the points (0,-2) and (3,0). Now try using an axial symmetry in the x-axis to find a second point on the required line. Draw a diagram to help picture things.

There are probably more sophisticated methods using the tan theta fornula from the log tables.

26-06-2012, 16:37	#1
fade out Registered User Join Date: May 2009 Posts: 23 Adverts \| Friends	Dividing a line in given ratio (brain freeze) Can someone help I can't figure out where I have gone wrong here. It is probably staring at me in the face but I have got brain freeze or something! A(4,-3) and B(-2,0) are the end points of a line segment. The point P(2,-2) divides [AB] internally in the ratio h:k. Find h:k. I have subbed in the co ordinates into the formula for internal division and then let each coordinate equal to a coordinate in P(2,-2). The equations I get from this however are no good as I seem to end up with -4h+2k=0 and 2h-k=0. Can someone point where I have gone wrong?

26-06-2012, 16:52	#4
fade out Registered User Join Date: May 2009 Posts: 23 Adverts \| Friends	Thanks invisible! Is there a method using the formula though? Also your method works too Delphi thank you! Last edited by fade out; 26-06-2012 at 17:01.

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26-06-2012, 16:43	#2
im invisible Closed Account Join Date: Feb 2011 Posts: 4,188 Adverts \| Friends	just take either the x or the y values, Y makes it easier A -3 P -2 B 0 so thats 1:2

26-06-2012, 16:52	#3
Delphi91 Registered User Join Date: May 2001 Location: Limerick Posts: 1,315 Adverts \| Friends	You could always use the "length of a line" formula to calculate \|AP\| and \|PB\|. Let H = \|AP\| and K=\|PB\|. Put them together in a ratio and simplify.

27-06-2012, 00:12	#5
MathsManiac Moderator Join Date: Apr 2007 Posts: 1,327 Adverts \| Friends Mod: Mathematics	You didn't go wrong. Either one of your two equations tells you that k=2h, so that tells you that h:k is 1:2, or 2:4, or 3:6, etc., which all represent the same ratio, as long as the second number is twice the first. (The fact that your two equations are equivalent just confirms that the given point is indeed on the line AB)

28-06-2012, 23:29	#6
fade out Registered User Join Date: May 2009 Posts: 23 Adverts \| Friends	Thanks maths maniac makes sense now! I'm onto another problem now! Find the equation of the line such that the x-axis bisects the angle between it and the line 2x-3y-6=0

28-06-2012, 23:55	#7
mikeystipey Registered User Join Date: Jul 2008 Location: The Pale / The Premier Posts: 179 Adverts \| Friends	Try and give us your attempt first OP - here's a starting point for one way to do it: Find 2 points on the line 2x-3y-6=0 by subbing in x=0 and y=0 respectively, giving the points (0,-2) and (3,0). Now try using an axial symmetry in the x-axis to find a second point on the required line. Draw a diagram to help picture things. There are probably more sophisticated methods using the tan theta fornula from the log tables.