Author Topic: solutions? (Read 6444 times) Tweet Share

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.

ice_blockie · « **Reply #1 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...

khalil · « **Reply #2 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.

khalil · « **Reply #3 on:** February 15, 2009, 11:44:59 am »

wait i dont understand!
how did u know the y value of R is 0?

Flaming_Arrow · « **Reply #4 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

$a +2 = 6-4a$

$\implies a = \boxed{\frac{4}{5}}$

khalil · « **Reply #5 on:** February 15, 2009, 03:29:29 pm »

thanks mate !

Mao · « **Reply #6 on:** February 15, 2009, 05:59:40 pm »

oops, misread the question

wombifat · « **Reply #7 on:** February 17, 2009, 07:38:41 pm »

theres a solution supplement on the cd? sweet as

khalil · « **Reply #8 on:** February 19, 2009, 03:44:38 pm »

yer but its like $50 for the CD

khalil · « **Reply #9 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

Flaming_Arrow · « **Reply #10 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

$(\frac{7}{2},\frac{-15}{4})$

khalil · « **Reply #11 on:** February 19, 2009, 07:18:50 pm »

thank you so much....do u have the essentials book too?

khalil · « **Reply #12 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

Flaming_Arrow · « **Reply #13 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

khalil · « **Reply #14 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))

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