prelim question (1 Viewer)

elliottjohnson · May 31, 2009

hey,
ive got a question from my mathematics extension assignment and ive need a little bit of help. thought it might be fun for those of u in year 12.

For the points A(-5,-1) and B(1,0):

(i) Write down the coordinates of P, the point that divided AB in the ratio K:1

(ii) If P lies on xy=1, show that k^2 + 3k - 4 = 0

(iii)Hence, or otherwise, find the coordinates of the points where the line AB meets xy=1.

thanks for any help.

jet · May 31, 2009

now, I didn't write out all the steps, because it is an assignment, and you need to do some work, but here is the basic outline:
$\text{For the division of an interval } k:1 \\ x = \frac{k - 5}{1 + k} \\ y = \frac{-1}{1 + k} \\ \\ \text{If P lies on } xy = 1 \\ \frac{k - 5}{1 + k} = -k - 1 \\ k - 5 = -(k^2 + 2k + 1) \\ k^2 + 3k - 4 = 0 \\ \\ \text{By solving this equation, } k = -4 $ or $ k = 1 \\ \text{Therefore, these values of k will solve the point P which lies on AB, and these solutions obey xy = 1 from the previous step} \therefore P \equiv (-2, \frac{-1}{2}) $ or $ P \equiv (3,\frac{1}{3})$

elliottjohnson · May 31, 2009

jetblack2007 said:
now, I didn't write out all the steps, because it is an assignment, and you need to do some work, but here is the basic outline:
$\text{For the division of an interval } k:1 \\ x = \frac{k - 5}{1 + k} \\ y = \frac{-1}{1 + k} \\ \\ \text{If P lies on } xy = 1 \\ \frac{k - 5}{1 + k} = -k - 1 \\ k - 5 = -(k^2 + 2k + 1) \\ k^2 + 3k - 4 = 0 \\ \\ \text{By solving this equation, } k = -4 $ or $ k = 1 \\ \text{Therefore, these values of k will solve the point P which lies on AB, and these solutions obey xy = 1 from the previous step} \therefore P \equiv (-2, \frac{-1}{2}) $ or $ P \equiv (3,\frac{1}{3})$

thanks

prelim question (1 Viewer)

elliottjohnson

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jet

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