2. ## can u see it this time?

If u can't read it, the question says "Supply the coordinates for each figure without introducing any new coordinates.

question (a) says: Triangle ABC has a right angle at A and angle B=Pie over 3.

question (b) says: BCi s a quarter circle with centre at (0,0). Line AD has equation y=x.

3. In part a, the slope of line BC is $\tan \left( {\frac{{2\pi }}{3}} \right)$.

4. Originally Posted by Raiden_11

If u can't read it, the question says "Supply the coordinates for each figure without introducing any new coordinates.

question (a) says: Triangle ABC has a right angle at A and angle B=Pie over 3.
The length of the base of the triangle is $a$.

Now, using the trig ratio for tangent, we see that

$\tan B = \frac {|AC|}{a}$ ........ $|AC|$ means the length of the line connecting A and C

$\Rightarrow \tan \left( \frac { \pi}{3} \right) = \frac {|AC|}{a}$

$\Rightarrow \sqrt {3} = \frac {|AC|}{a}$

$\Rightarrow a = \frac {|AC|}{ \sqrt {3}} = \frac { \sqrt {3} |AC|}{3}$

So we have $B \left( \frac { \sqrt {3} |AC|}{3}, 0 \right)$

question (b) says: BCi s a quarter circle with centre at (0,0). Line AD has equation y=x.
Note that $r = |AD| = |AB|$

Also note that the angle the line y = x makes with the x-axis is $45^{ \circ}$, or $\frac { \pi}{4}$

Using the trig ratio for sine we see that

$\sin \left( \frac { \pi}{4} \right) = \frac {|DE|}{|AD|} = \frac {|DE|}{r}$ .........E is the point where a vertical line from D cuts the x-axis, |DE| represents the length of that line

$\Rightarrow r = \frac {|DE|}{ \frac { \sqrt {2}}{2}} = \frac {2 |DE|}{ \sqrt {2}}$

So $B \left( 0, \frac {2 |DE|}{ \sqrt {2}} \right)$

These questions are a bit vague to me. I guess we could do the last one without introducing a new point, if we used the cosine rule or something like that

5. It appears to me as if the question (a) is asking for the coordinates of C.
Because line BC has slope $\tan \left( {\frac{{2\pi }}{3}} \right) = - \sqrt 3$ then say C(0,c) so $\frac{{c - 0}}{{0 - a}} = - \sqrt 3 \quad \Rightarrow \quad c = a\sqrt 3 \quad or\quad C:\left( {0,a\sqrt 3 } \right).$
Thus we have written the coordinates of C in terms of the given.

6. Originally Posted by Plato
It appears to me as if the question (a) is asking for the coordinates of C.
Because line BC has slope $\tan \left( {\frac{{2\pi }}{3}} \right) = - \sqrt 3$ then say C(0,c) so $\frac{{c - 0}}{{0 - a}} = - \sqrt 3 \quad \Rightarrow \quad c = a\sqrt 3 \quad or\quad C:\left( {0,a\sqrt 3 } \right).$
Thus we have written the coordinates of C in terms of the given.
oh yeah, silly me. and i suppose question (b) is asking for coordinate D right...or C? or both, yeah, both

well, i'll leave my post up there, just in case anybody wonders, "well, how does a and r relate to all of this?"

7. ## Here's another question guys....it seems simple but i think theres more to it.

Thanks for the help earlier can u help me answer this?

8. ## I really need help!

These are questions from the grade 12 geometry course i am taking.

1. Triangle ABC has medians CE and BD as shown in the diagram. Use a coordinate proof and a vector proof to prove BF : FD = 2 : 1

9. Coordinates prove:
We can assume that $A\in Oy,B,C\in Ox$ and $A(0,2a),B(2b,0),C(2c,0)$, for the simplicity of calculus.
Now $D(c,a),E(b,a)$. Then you can write the equations of $BD$ and $CE$, solve the system and you will find the coordinates of the point $F$. Then you can calculate the length of the segments $BF,FD$.

10. ## thanks....but

Thanks but can you show me how to solve it just so i can have a clear idea.

11. ## Also

Also,how did you get those points in the first place...can u show me in detail please. Much appreciated.

12. Given a triangle $ABC$ in plane we can get the $Ox$ and $Oy$ axis anywhere in that plane. So we can get $Ox$ axis passing through $B$ and $C$ and $Oy$ axis passing through $A$.
Now, if there are two points $M_1(x_1,y_1),M_2(x_2,y_2)$, then the midpoint $M$ of the segment $M_1M_2$ has the coordinates $\displaystyle \frac{x_1+x_2}{2}$ and $\displaystyle \frac{y_1+y_2}{2}$. Then $D(c,a), \ E(b,a)$.
The equation of the line passing through $M_1(x_1,y_1),M_2(x_2,y_2)$ is $\displaystyle \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$ or
$\left|\begin{array}{lll}
x & y & 1\\
x_1 & y_1 & 1\\
x_2 & y_2 & 1
\end{array}\right|=0$

Using one of these equations you can now write the equations of $BD$ and $CE$, then solve the system formed by these equations to find the coordinates of $F$.
Then you can use the distance formula between two points in plane:
$M_1M_2=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
to calculate $BF$ and $FD$
It is more clear now?

13. Vector prove:
Let $\displaystyle \frac{BF}{FD}=k$
Then $\displaystyle \overrightarrow{CF}=\frac{1}{1+k}\overrightarrow{C B}+\frac{k}{1+k}\overrightarrow{CD}$ and
$\displaystyle \overrightarrow{CE}=\frac{1}{2}(\overrightarrow{CB }+\overrightarrow{CA})=\displaystyle \frac{1}{2}\left(\overrightarrow{CB}+2\overrightar row{CD}\right)=\frac{1}{2}\overrightarrow{CB}+\ove rrightarrow{CD}$.
The vectors $\overrightarrow{CF}$ and $\overrightarrow{CE}$ are colinear, so
$\displaystyle \frac{\frac{1}{1+k}}{\frac{1}{2}}=\frac{\frac{k}{1 +k}}{1}\Rightarrow k=2$, so $\displaystyle \frac{BF}{FD}=2$

14. ## I have another problem

Thanks man, i really appreciate the help but i have two more problems.