# Thread: looks like coordinate geometry but is not really...

1. ## looks like coordinate geometry but is not really...

The points A and B have coordinates (3,-1) and (6,3) respectively. The points C and D are each distant 4 units from A and 6 units from B, forming a kite. Show that the length of CD is 3 x root of 7 (i dont get how you people manage to type math symbols).
I figured that this is supposed to be dealed with using the cosine rule and not by trying to calculate the coordinates of C and D, which would take ages. I managed to get the right answer, which is 7.blablablabla, but not in the right form, that is in decimal form instead of surd form. I really can't figure out how to get a direct surd.

2. Originally Posted by the kopite
The points A and B have coordinates (3,-1) and (6,3) respectively. The points C and D are each distant 4 units from A and 6 units from B, forming a kite. Show that the length of CD is 3 x root of 7 (i dont get how you people manage to type math symbols).
I figured that this is supposed to be dealed with using the cosine rule and not by trying to calculate the coordinates of C and D, which would take ages. I managed to get the right answer, which is 7.blablablabla, but not in the right form, that is in decimal form instead of surd form. I really can't figure out how to get a direct surd.
Have you tried drawing everything on a graphic, then setting up the points C and D by making sure you follow the given instructions? You can find the coordinates of C & D by simply drawing the kite, then you could use $\displaystyle d=\sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2)}$ to find the distance between both points. Plus, finding points using that equation will also allow you to keep your answer in a direct surd (in this case, a square root).

3. Hello the kopite
Originally Posted by the kopite
The points A and B have coordinates (3,-1) and (6,3) respectively. The points C and D are each distant 4 units from A and 6 units from B, forming a kite. Show that the length of CD is 3 x root of 7 (i dont get how you people manage to type math symbols).
I figured that this is supposed to be dealed with using the cosine rule and not by trying to calculate the coordinates of C and D, which would take ages. I managed to get the right answer, which is 7.blablablabla, but not in the right form, that is in decimal form instead of surd form. I really can't figure out how to get a direct surd.
Do you know the formula for the area of a triangle if you know the lengths of all three of its sides? Read on!

$\displaystyle AB^2 = 3^2+4^2 = 25$

$\displaystyle \Rightarrow AB = 5$

By symmetry, $\displaystyle CD$ and $\displaystyle AB$ are perpendicular, so in $\displaystyle \triangle ABC$, with $\displaystyle AB$ as base, the height of the triangle, $\displaystyle h = \tfrac12CD$.

Now $\displaystyle AB =5, AC = 4, CD = 6$, so the semi-perimeter, $\displaystyle s = \tfrac12(4+5+6) = \frac{15}{2}$.

Area of $\displaystyle \triangle ABC = \sqrt{s(s-a)(s-b)(s-c)} = \frac{15\sqrt7}{4} = \tfrac12.5.h$ (Did you know that formula?)

$\displaystyle \Rightarrow h = \frac{3\sqrt7}{2}$

$\displaystyle \Rightarrow CD = 3\sqrt7$.

4. ummmm...first of all i have not been taught that formula so i'm sure there's another way. also, you confuse me by writing at a certain point that CD=6...What?

5. Originally Posted by scouser
ummmm...first of all i have not been taught that formula so i'm sure there's another way. also, you confuse me by writing at a certain point that CD=6...What?
Originally Posted by the kopite
but i agree, your expalnation is way too ambiguous, grandad, and im not sure about that formula either...
Well, if you had drawn the figure, you would have immediately realized that it was a simple typo. Grandad meant BC = 6 cm

6. ok thanks, that clears things up. i just didnt know the math wizards made typos...

7. ## Alternative to 's' formula

Originally Posted by the kopite
ok thanks, that clears things up. i just didnt know the math wizards made typos...
'Fraid so!

PS

If you want an alternative that doesn't use the 's' formula, how about this?

Suppose AB and CD meet at E, and the length $\displaystyle AE = x$. Then $\displaystyle EB = 5-x$.

Then there are two right-angled triangles, AEC and ECB, in which we can write an expression for $\displaystyle h$ (the distance CE). Using these, we get:

$\displaystyle h^2 = 4^2 - x^2 = 6^2 - (5-x)^2$

$\displaystyle \Rightarrow 16-x^2 = 36-25+10x-x^2$

$\displaystyle \Rightarrow 10x = 5$

$\displaystyle \Rightarrow x = \tfrac12$

$\displaystyle \Rightarrow h^2 = 16 - \tfrac14 = \frac{63}{4}$

$\displaystyle \Rightarrow h = \frac{3\sqrt7}{2}$

(Mind you, I think you should learn the 's' formula - it comes in useful!)