# 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 $d=\sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2)}$ 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!

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

$\Rightarrow AB = 5$

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

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

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

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

$\Rightarrow CD = 3\sqrt7$.

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

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

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

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

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

$\Rightarrow 10x = 5$

$\Rightarrow x = \tfrac12$

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

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

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