O-Level Additional Math

[Guan Hui]

I meant this thread for all secondary questions

[schellen]

Yes, but since it'll probably get messy, and there are existing threads for E and A Maths, perhaps, you can start level specific threads instead?

We can sticky them for you.

[tankee]

do we really need separate threads for each level of Mathematics?


Wouldn’t just 1 maths thread suffice?

[tianzhu]

Hi schellen and tankee


I believe Guan Hui is a tutor.I think he is trying to start a dedicated thread similar to that of mathsguru in the primary section.

Maybe the moderator can consider to give him a sticky thread for him to help to solve Secondary Maths problems for members. This is a win-win situation, members can get help from a professional tutor and GH can gain more exposure about his tutoring service.Let’s us take a small step first and monitor the responses, questions for all secondary levels can be lumped together.But members asking questions should state the level of their question.

Perhaps, the thread’s heading can be
Tutor Guan Hui ------ Ask me your Secondary Maths Questions

As for GH, I hope you are mentally prepared for the long journey.It takes a lot of stamina to do what mathsguru is doing.

Best wishes

[Guan Hui]

Thx tianzhu it is exactly what i intended. and thankyou for your wishes

[schellen]

Then, go ahead and edit this thread's title, Guan Hui...and good luck!

[tianzhu]

Tutor Guanhui- Ask Your Secondary Questions Here!


Hi GH

What are the subjects?

Best wishes

[Guan Hui]

oppps... haha thanks for reminding

[tweet]

Hi Guan Hui


Pls help with these 2 questions:

1) Given that A is the point (0,2), B is the point (9,0) and C is a point on the line y=2x + 2 such that AB = BC, find the coordinates of C. If ABCD forms a parallelogram, find the coordinates of point D.
Answers: C(2,6) and D(-7,8 )

2) Find the area of triangle ABC whose coordinates are A(3,3), B(-1,0) and C (5,-3) Hence, or otherwise, calculate the distance from C to AB.
Answer: 15;6

TIA

[Guan Hui]

zzz… typing this for the 3rd time… haha here we go


let c be (x,y)
distance of ab=sqrt(9^2+2^2)
=sqrt(85)
distance of bc=sqrt((9-x)^2+y^2)
since ab=bc
sqrt((9-x)^2+y^2)=sqrt(85)
(9-x)^2+y^2=85
y^2=85-81+18x-x^2
y^2=4+18x-x^2------------------1
y=2x+2
y^2=(2x+2)^2
y^2=4x^2+8x+4------------------2
4+18x-x^2=4x^2+8x+4
5x^2-10x=0
x=0 (rejected cause this value is for a)
or x=2
put x=2 into y=2x+2
y=6
c(2,6)

since abcd is parallelogram, ab ll cd and bc ll ad
gradient ab=(2-0)/(0-9)
=-2/9
y =-2/9x +c
sub c in,
6=-4/9+c
c=58/9
equation of cd: y=-2/9x+58/9

gradient of bc=(0-6)/(9-2)
=-6/7
equation of line ad: y= -6/7x +2

-2/9x+58/9=-6/7x+2
-14x+406=-54x+126
40x=-280
x=-7
put x=-7 into y=-6/7x+2
y=8

d(-7,8 )