Question

Coordinate geometry problems x2

Answer 1

These are the nice ones! Enjoy!~

Answer 2

A=(-6, 0)
B=(0, -6)

Answer 3

Is AB the diameter in #1?

anyways, A=(-6,0), B=(0,-6)

Answer 4

Yes.

Answer 5

yeah ofcourse its the diameter, that AB. #circumcircle

Then you have the length of the diameter from distance formula. also midpoint of the diameter shall serve as the centre. get that through the midpoint (section) formula. problem is Solved!!

Equation of circle+

\[(x+3)^2 + (y+3)^2 = 18\]

Answer 6

For 5b(ii).
\[
\begin{align*}
\text{If }AC^2+CO^2=AO^2\text{, then }\angle ACO=90^\circ. \\
\text{By 5a, }A=(-6,0),\,B=(0,-6),\,C=(-3,-3) \\
AB=BO=\sqrt{18} \\
AO=6 \\
AB^2+BO^2&=2(\sqrt{18}^2) \\
AB^2 +BO^2&=36 \\
\because AB^2+BO^2=AO^2 \\
\therefore \angle ACO=90^\circ
\end{align*}
\]

Answer 7

For 5b(ii), please help me to complete the following as I am using the slow and useless iPad 2.
1. Calculate the equation for CO and AF.
2. Calculate the coordinates of the intersection CO and AF (i.e. point K).
3. Calculate the angle OKF by using the cosine law.

Answer 8

5 b) iii) <ACF = 135 <--- rest is easy.

Answer 9

5 b)) 90

Answer 10

6) 1