Author Topic: Maths Quest!!!! (Read 2189 times) Tweet Share

Conic · « **Reply #15 on:** April 25, 2014, 11:22:15 pm »

They answered that question in another thread recently: it's so hard :(

bts · « **Reply #16 on:** April 25, 2014, 11:34:41 pm »

yea... i know... but i still couldn't ans it cause before P lies on a straight line so you knew one of the coordinates already being 7, but with this one the P is on a slanting line so i'm not sure about any of the co ordinates making it hard... maybe its just me?

?

Conic · « **Reply #17 on:** April 26, 2014, 01:48:11 pm »

So you let P=(x,y). Now

$\overrightarrow{ZP} = \overrightarrow{OP} - \overrightarrow{OZ} = x \underset{\sim}{i} + y \underset{\sim}{j} -5\underset{\sim}{i} = (x-5)\underset{\sim}{i}+y \underset{\sim}{j}$

We know that $\overrightarrow{OY} = \overrightarrow{OX} + \overrightarrow{OZ} = 7 \underset{\sim}{i} + 7 \underset{\sim}{j}$ , and we know that ZP and OY are perpendicular, so
$\overrightarrow{ZP} \cdot \overrightarrow{OY} = 0$

$\implies 7\times (x-5)+7 \times y = 0$

$\implies x = 5-y$

We also know that y=7, so x=5-7=-2.

bts · « **Reply #18 on:** April 26, 2014, 02:10:56 pm »

but i think the ans says (12 and 3/8, 10 and 5/8)

i think the x = -2 is for the other question:

What i did was:
Let P (xi + yj)
XP . XZ = 0
XP = (-3+x)i +(-5+y)j
XZ = (3i - 5j)
[(-3+x)i +(-5+y)j] . (3i - 5j) = 0
-9 + 3x +25 -5y = 0 (1)

and the 1MIN said:
PZ = k YZ (cause it needs to be colinear to it?) but then i'm introducing another variable and another equation is needed if i have to work it out simultaneously..

Conic · « **Reply #19 on:** April 26, 2014, 02:53:55 pm »

Looking at this: Re: it's so hard :( it doesn't look like XP and XZ are perpendicular, so their dot product isn't 0. Also, you only have one variable, since you know y=7.

bts · « **Reply #20 on:** April 26, 2014, 02:58:40 pm »

oh that one is a different question to the one i'm trying to solve. That one is like 3 something and the one i'm looking at is 6g) i believe. They are different questions
They are both similar and i've tried to apply what they did to this question but because before P lies on a straight line so you knew one of the coordinates already being 7, but with this one the P is on a slanting line so i'm not sure about any of the co ordinates making it hard.

the question was posted here: sigh... vectors!!!

Conic · « **Reply #21 on:** April 26, 2014, 04:00:27 pm »

Yeah my bad, I was looking at the wrong question.

For that question you can use linear graphs.

The gradient of XZ is (5-0)/(3-6)=-5/3, to the gradient of a line perpendicular to XZ is -1/(-5/3)=3/5. The line XP has the gradient 3/5 (because it is perpendicular to XZ), and it passes through (3,5), so it's equation is y=3/5x+16/5.

The gradient of ZY is (5-0)/(9-6)=5/3, so the gradient of ZP is 5/3 (because P is on the line ZY). This line passes through (6,0), so it has the equation y=5/3x-10.

P is on the intersection of these lines, so to find the x coordinate we have 3/5x+16/5=5/3x-10, and find this x and substitute it into one of the equations to find y.

This gives x=99/8=12+3/8, and y=85/8=10+5/8.

sujui · « **Reply #22 on:** April 26, 2014, 04:10:05 pm »

dear conic,

you're helping alot on this thread, I was wondering can you help prove how the medians of a triangle are concurrent? I know concurrent means that they intersect at one point only...

Conic · « **Reply #23 on:** April 26, 2014, 05:07:28 pm »

I think this proves it:

For some triangle OAB, let the M be the midpoint of OA, let N be the midpoint of OB, and let X be the midpoint of AB. Let $\overrightarrow{OA}=\underset{\sim}{a}$ and $\overrightarrow{OB}=\underset{\sim}{b}$ . Let P be the point where the medians AN and BM intersect. We will show that O, P and X are collinear, which means that P is a point on the median OX that is also the point of intersection of the other two medians, which means the three medians meet at P i.e., they are concurrent.

AP is parallel to AN, so $\overrightarrow{AP}=\lambda \overrightarrow{AN}$ for some real number $\lambda$ . $\overrightarrow{AN}=\frac{1}{2}\underset{\sim}{a}-\underset{\sim}{b}$ , so we have $\overrightarrow{AP}=\lambda (\frac{1}{2}\underset{\sim}{b}-\underset{\sim}{a})$ .

BP is parallel to BM, so $\overrightarrow{BP}=\mu \overrightarrow{BM}$ for some real number $\mu$ . $\overrightarrow{BM}=\frac{1}{2}\underset{\sim}{a}-\underset{\sim}{b}$ , so we have $\overrightarrow{BP}=\mu(\frac{1}{2}\underset{\sim}{a}-\underset{\sim}{b})$ .

$\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}=\underset{\sim}{a}+\lambda(\frac{1}{2}\underset{\sim}{b}-\underset{\sim}{a})=(1-\lambda)\underset{\sim}{a}+\frac{\lambda}{2}\underset{\sim}{b}$

$\overrightarrow{OP}=\overrightarrow{OB}+\overrightarrow{BP}=\underset{\sim}{b}+\mu(\frac{1}{2}\underset{\sim}{a}-\underset{\sim}{b})=\frac{\mu}{2}\underset{\sim}{a}+(1-\mu)\underset{\sim}{b}$

$\implies (1-\lambda)\underset{\sim}{a}+\frac{\lambda}{2}\underset{\sim}{b}= \frac{\mu}{2}\underset{\sim}{a}+(1-\mu)\underset{\sim}{b}$

Since $\underset{\sim}{a}$ and $\underset{\sim}{b}$ are linearly independent we have $1-\lambda = \frac{\mu}{2}$ and $1-\mu=\frac{\lambda}{2}$ , which gives $\mu=\lambda=\frac{2}{3}$ .

Now we know that $\overrightarrow{OP}=(1-\frac{2}{3})\underset{\sim}{a}+\frac{\frac{2}{3}}{2}\underset{\sim}{b}=\frac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b})$ .

$\overrightarrow{OX}=\overrightarrow{OA}+\frac{1}{2} \overrightarrow{AB}=\underset{\sim}{a}+\frac{1}{2}(\underset{\sim}{b}-\underset{\sim}{a})=\frac{1}{2}(\underset{\sim}{a}+\underset{\sim}{b})$ .

Therefore O, P and X are collinear, as OP and OX are parallel and they share the point O. This means that P lies on the median OX. Since P is the point of intersection of the other two medians and it is on the median OX, all three medians meet at P, i.e., they are concurrent.

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Author Topic: Maths Quest!!!! (Read 2189 times) Tweet Share

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