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Confusing Straight Line Q - HSN forum

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Confusing Straight Line Q


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#1 DX_Suck_It

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Posted 05 September 2006 - 05:28 PM

*ABCD is a square with A the point (2,3) and C(6,5). Find the point of intersection of the diagonals and the co-ords of B and D.


I can't get it, so could somene explain to me how I can solve this question and get the answer, thanks.

#2 DX_Suck_It

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Posted 05 September 2006 - 08:47 PM

C'mon, could somebody please help me out here.

#3 The Wedge Effect

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Posted 05 September 2006 - 09:01 PM

I haven't done Higher maths for over a year now, so my memory's a little sketchy, but what you could do to find the point of intersection is use the mid-point formula, as the point of intersection would be at the middle of the two points, as the question states it's a perfect square. You should, as a result, get (4,4) as your answer. However, as I can't remember the distance formula, I'll leave the next part to someone who can, unless there's another way, which I hadn't noticed. God, I am terrible...starting maths at uni, and not remembering something this simple. laugh.gif

#4 George

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Posted 05 September 2006 - 10:13 PM

Well, I got an answer, but it was quite involved. I'm maybe not seeing the easiest way, mind you.

Let M be the point where the diagonals meet, so M(4,4).

The diagonal AC has gradient 1/2, so the other diagonal (on which B and D lie) has gradient -2. Since this passes through M, we can write down the equation of the other diagonal:



\begin{align*}y-4&=-2(x-4)\\ y&=12-2x\end{align*}

We have dAM2 = 5 (you can check this), and since ABCD is a square, it follows that dBM2 = dDM2 = 5.

Using the point (x,12-2x) which lies on the other diagonal, this point will be one of B or D if it satisfies



\begin{align*}(x-4)^2+(12-2x-4)^2=5\end{align*}

(using the distance formula)

Solving this gives two solutions, and these x-values can be put into (x,12-2x) to give the coordinates of B and D.

I suspect there is maybe something missing from the question as posted, because that is quite a long path to the answer. Either that, or I'm just doing it the hard way smile.gif





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