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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: khalil on February 15, 2009, 09:06:28 am

Title: solutions?
Post by: khalil on February 15, 2009, 09:06:28 am: Someone help please

The point (h,k) lies on the line y=x+1 and is 5 units from the point (0,2). Write down two equations connecting h and k and hence find possible values of h and k.

This question is from essential's.
Does any one have the solutions supplement on CD? Please share, the questions here are relli hard.
Title: Re: solutions?
Post by: ice_blockie on February 15, 2009, 09:59:27 am: Since $(h, k)$ lies on $y= x+1$ therefore $k = h + 1$

Also, using distance formula: $\sqrt{(2-k)^2+(0-h)^2}=5$

Using simultaneous equations, you get the answer.

And no, I don't have the solutions...
Title: Re: solutions?
Post by: khalil on February 15, 2009, 10:29:37 am: thanks i get it now....wat about this one:

P and Q are the points of intersection of the line y\2+x\2=1 with the x and y axes respectively. The gradient of QR is 1\2, where R is the point with x-coordinate 2a, a>0.
a) Find the y-coordinate of R in terms of a
b) Find the value of 'a' if the gradient of PR is -2.
Title: Re: solutions?
Post by: khalil on February 15, 2009, 11:44:59 am: wait i dont understand!
how did u know the y value of R is 0?
Title: Re: solutions?
Post by: Flaming_Arrow on February 15, 2009, 11:48:51 am: a. $\frac {y}{2} + \frac{x}{3} = 1$

gradient of QR $\frac{1}{2}$ where $R(2a,y)$ where $a > 0$

$P(3,0)$ and $Q(0,2)$

use gradient formula to get the y coordinate of R

$\frac {2-y}{0-2a} = \frac{1}{2}$

$\implies y = \boxed{ a +2}$

b. gradient of PR = -2

$\frac{0-y}{3-2a} = -2$

$y = 6-4a$

then we have our two equations to find the value of a

$a +2 = 6-4a$

$\implies a = \boxed{\frac{4}{5}}$
Title: Re: solutions?
Post by: khalil on February 15, 2009, 03:29:29 pm: thanks mate !
Title: Re: solutions?
Post by: Mao on February 15, 2009, 05:59:40 pm: oops, misread the question
Title: Re: solutions?
Post by: wombifat on February 17, 2009, 07:38:41 pm: theres a solution supplement on the cd? sweet as
Title: Re: solutions?
Post by: khalil on February 19, 2009, 03:44:38 pm: yer but its like $50 for the CD
Title: Re: solutions?
Post by: khalil on February 19, 2009, 05:56:49 pm: If A = (-4,6) and B = (6,-7), find;
the coordinates of P, where P is an element of AB and AP:PB=3:1
I hate essentials, i hope the exams are not that hard
Title: Re: solutions?
Post by: Flaming_Arrow on February 19, 2009, 06:12:52 pm: $\frac{AP}{PB}=\frac{3}{1}$

$AP = 3PB$

$AB = AP + PB$

$AB = 4PB$

$\frac{AB}{4} = PB$

find the magnitude of AB

$\sqrt{(-7-6)^2+(6+4)^2}= \sqrt{269}$

$\frac{\sqrt{269}}{4} = PB$

$\frac{3\sqrt{269}}{4} = AP$

then

$\frac{\sqrt{269}}{4} = \sqrt{(6-x)^2+(-7-y)^2}$

$\frac{3\sqrt{269}}{4} = \sqrt{(-4-x)^2+(6-y)^2}$

then using the CAS calculator work out x and y

$(\frac{7}{2},\frac{-15}{4})$
Title: Re: solutions?
Post by: khalil on February 19, 2009, 07:18:50 pm: thank you so much....do u have the essentials book too?
Title: Re: solutions?
Post by: khalil on February 19, 2009, 08:05:16 pm: what about this one?
It is very similar
the coordinates of P, where P is an element of AB and AP:AB=3:1
Title: Re: solutions?
Post by: Flaming_Arrow on February 19, 2009, 09:07:11 pm: draw the points on line

AP=3AB

AP-AB = BP

AB = $\sqrt{256}$

AP = $3 \sqrt{256}$

BP = $2 \sqrt{256}$

i think you can do it from there
Title: Re: solutions?
Post by: khalil on February 22, 2009, 07:56:25 pm: yer i can....thanks
wat about this one?
show that (3x+2)/(x+1)= 3-(1/(x+1))
Title: Re: solutions?
Post by: TrueTears on February 22, 2009, 08:04:01 pm: have u tried polynomial long division?
Title: Re: solutions?
Post by: Flaming_Arrow on February 22, 2009, 08:08:55 pm: work backwards

$3-\frac{1}{x+1}$

$\frac{3(x+1)-1}{x+1}$

$\frac{3x+3-1}{x+1}$

$\frac{3x+2}{x+1}$

LHS=RHS