Cordinate geometry

**Coach** · March 9th 2008, 01:49 AM

Dear forum members,

could you please check if my calculations are correct, I have tried several times, I alway keep getting the wrong answer though.

Problem

Find the values for k for which the lines $2x+ky-11=0$ and $(3-k)x+y+2=0$ are perpendicular.

My solution

$-3x+kx= \frac{k}{2x}$
-3x+kx and 2x have to be the same

so $-3x+kx= 2x$

I always keep getting 5 or 1 as the answer for k, is it correct?

Thank you in advance!

**mr fantastic** · March 9th 2008, 01:54 AM

Originally Posted by Coach

Dear forum memebers,

could you please check if my calculations are correct, I have tried several times, I alway keep getting the wrong answer though.

Problem

Find the values for k for which the lines $2x+ky-11=0$ and $(3-k)x+y+2=0$ are perpendicular.

My solution

$-3x+kx= //frac{k}{2x}$
the bottom numbers have to be the same

so $-3x+kx= 2x$

I always keep getting 5 or 1 as the anwer for k, is it correct?

Thank you in advance!

Two lines are perpendicular if the product of their gradients is equal to -1.

$2x+ky-11=0 \Rightarrow y = \frac{-2x+11}{k} \Rightarrow m = -\frac{2}{k}$ .

$(3-k)x+y+2=0 \Rightarrow y = -(3 - k)x - 2 = (k - 3)x - 2 \Rightarrow m = k - 3$ .

Therefore you require $\left( -\frac{2}{k} \right) \, (k-3) = -1 \Rightarrow -2(k-3) = -k \Rightarrow k = 6$ .

**Coach** · March 9th 2008, 01:59 AM

Thank you so much!

I would not have been able to do that on my own.

**mr fantastic** · March 9th 2008, 03:02 AM

Originally Posted by Coach

Dear forum members,

could you please check if my calculations are correct, I have tried several times, I alway keep getting the wrong answer though.

Problem

Find the values for k for which the lines $2x+ky-11=0$ and $(3-k)x+y+2=0$ are perpendicular.

I always keep getting 5 or 1 as the answer for k, is it correct?

Thank you in advance!

I just realised that if you kept getting 5 plus 1 as the answer for k it would be correct

**mr fantastic** · March 9th 2008, 03:06 AM

Originally Posted by Coach

Thank you so much!

I would not have been able to do that on my own.

I'm sure you would

You've just got to work through each link in the chain.

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