area of quadrilateral

**afeasfaerw23231233** · June 14th 2008, 03:54 AM

bk2 p105 q27
question : the vertices of a quadrilateral are the centres of the circles:
$C_1 : x^2 +y^2+2tx=0$
$C_2: x^2+y^2 +\frac {2y}t = 0$
and their intersecting points
a) find the coordinates of the vertices of the quadrilaterl.
b) find that the area of the quadrilateral is a constant.
my working
vertices :
-t, 0
$0,- \frac 1 t$
0,0
$-\frac {2t}{1+t^4} , -\frac {2t^3 }{1+t^4}$
area:
$\frac 1 2 \begin{vmatrix} 0 & 0 \\ -t & 0 \\ 0 & -\frac 1 t \\ -\frac {2t}{1+t^4} & -\frac {2t^3 }{1+t^4}\\ 0 & 0 \end{vmatrix}$
$= \frac 1 2 (\frac {t^4-1}{1+t^4})$
cannot prove that the area is a constant
thanks

**Isomorphism** · June 14th 2008, 04:17 AM

Originally Posted by afeasfaerw23231233

bk2 p105 q27
question : the vertices of a quadrilateral are the centres of the circles:
$C_1 : x^2 +y^2+2tx=0$
$C_2: x^2+y^2 +\frac {2y}t = 0$
and their intersecting points
a) find the coordinates of the vertices of the quadrilaterl.
b) find that the area of the quadrilateral is a constant.
my working
vertices :
-t, 0
$0,- \frac 1 t$
0,0
$-\frac {2t}{1+t^4} , -\frac {2t^3 }{1+t^4}$

Yes the co-ordinates are right.

But the area is wrong. The area is a constant and it is 1.

I did it the long way. First let $O_1O_2$ be the line joining the centers of $C_1$ and $C_2$ .

Then equation of the line $O_1O_2$ is $\frac{x}{t} + yt + 1 = 0$ .

Now the base of the triangle is $|O_1O_2| = \sqrt{t^2 + \frac1{t^2}}$ .

The perpendicular distance(height) from $(0,0)$ to the line is $\frac{|\frac{0}{t} + 0t + 1|}{\sqrt{t^2 + \frac1{t^2}}}$ . So the area of the triangle is $\frac12 |O_1O_2|\frac{|\frac{0}{t} + 0t + 1|}{\sqrt{t^2 + \frac1{t^2}}} = \frac12 |\frac{0}{t} + 0t + 1|= \frac12$ .

The perpendicular distance(height) from $(-\frac {2t}{1+t^4} , -\frac {2t^3 }{1+t^4})$ to the line is $\frac{|\frac{-\frac {2t}{1+t^4}}{t} -\frac {2t^3 }{1+t^4}t + 1|}{\sqrt{t^2 + \frac1{t^2}}}$ . So the area of the triangle is $\frac12 |O_1O_2|\frac{|-\frac {2}{1+t^4} -\frac {2t^4 }{1+t^4} + 1|}{\sqrt{t^2 + \frac1{t^2}}} = \frac12 |-\frac {2+2t^4 }{1+t^4} + 1| =\frac12 |-2 + 1| = \frac12$ .

So the area of the quadrilateral is the sum of the area of triangles.

And that is 1, a constant.

**mr fantastic** · June 14th 2008, 04:22 AM

Originally Posted by afeasfaerw23231233

bk2 p105 q27
question : the vertices of a quadrilateral are the centres of the circles:
$C_1 : x^2 +y^2+2tx=0$
$C_2: x^2+y^2 +\frac {2y}t = 0$
and their intersecting points
a) find the coordinates of the vertices of the quadrilaterl.
b) find that the area of the quadrilateral is a constant.
my working
vertices :
-t, 0
$0,- \frac 1 t$
0,0
$-\frac {2t}{1+t^4} , -\frac {2t^3 }{1+t^4}$
area:
$\frac 1 2 \begin{vmatrix} 0 & 0 \\ -t & 0 \\ 0 & -\frac 1 t \\ -\frac {2t}{1+t^4} & -\frac {2t^3 }{1+t^4}\\ 0 & 0 \end{vmatrix}$
$= \frac 1 2 (\frac {t^4-1}{1+t^4})$
cannot prove that the area is a constant
thanks

When t = 1, A = 1. But your answer gives A = 0 ......

**mr fantastic** · June 14th 2008, 04:43 AM

Originally Posted by mr fantastic

When t = 1, A = 1. But your answer gives A = 0 ......

$A = \frac{1}{2} \, |(x_1 y_2 - x_2 y_1) + (x_2 y_3 - x_3 y_2) + (x_3 y_4 - x_4 y_3) + (x_4 y_1 - x_1 y_4)|$

where the order of the points is such that 1 connects to 2 connects to 3 connects to 4 connects to 1.

So your answer is wrong because you have the order of the points wrong in the formula you used ......

You should get $A = \frac{1}{2} \left( \frac{2 t^4}{1 + t^4} + \frac{2}{1 + t^4}\right) = 1$ , as expected from the special case I considered in my earlier reply.

**earboth** · June 14th 2008, 05:15 AM

Originally Posted by afeasfaerw23231233

bk2 p105 q27
question : the vertices of a quadrilateral are the centres of the circles:
$C_1 : x^2 +y^2+2tx=0$
$C_2: x^2+y^2 +\frac {2y}t = 0$
and their intersecting points
a) ...
b) find that the area of the quadrilateral is a constant.
...

Here is a completely different attempt:

1. Calculate the coordinates of the centers of the 2 circles:

$x^2+2tx+{\color{red}t^2}+y^2 = {\color{red}t^2} ~\iff~ (x+t)^2+y^2 = t^2$

$x^2+y^2 +\frac {2y}t +{\color{red} \left(\frac1t \right)^2}={\color{red} \left(\frac1t \right)^2}~\iff~ x^2+\left(y+\frac1t\right)^2= \left(\frac1t \right)^2$

Then the area of the quadrilateral consists of 2 congruent right triangles. The legs of one of these right triangles have the lengthes(?) $R = t$ and $r = \frac1t$ .

Therefore the area of the quadrilateral is:

$A = 2 \cdot \frac12 \cdot t \cdot \frac1t = 1$

**afeasfaerw23231233** · June 14th 2008, 07:06 AM

i joined the points in a wrong order.
thanks for all your replies!

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