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please explain q.22 and q.23 

https://internal.challoners.com/x/departments/maths/gcse/solutions/edexcel-set-1/solutions-p1.pdf

for question 23 i struggle with understanding why the equation is y= -1/2x + c 

for question 22 is why CX = CA + AX also the rest is too confusting

Guest Oct 15, 2017
edited by Guest  Oct 15, 2017
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6+0 Answers

 #1
avatar+78755 
+2

For question 23 i struggle with understanding why the equation is y= -1/2x + c 

 

A = (-2, 0)    B = (0,4)   the slope of this line is  2

 

So....a perpendicular line to this will have the slope  =  -1/2

 

And since this line passes through  (5, -1) we have that

 

y = (-1/2)(x - 5) - 1

 

y =  (-1/2)x + 5/2 - 1

 

y = (-1/2)x + 3/2

 

The "c"  comes from the standard form of a line.....

Ax + By   = C

Subtract Ax from both sides

By  =  -Ax + C     divide both sides by B

y =  (-A/B) x  +  C/B 

 

Using what we know      -A/B  = the slope =  -1/2     y = -1    and x = 5

-1  = (-1/2) (5)  +  C/B

3/2  = C/B      where C/B   =  "c"

 

 

cool cool cool

CPhill  Oct 16, 2017
 #2
avatar+91053 
0

Here is question 22.

The question, not the answer :))

 

Melody  Oct 16, 2017
 #3
avatar+91053 
0

When I thought a and b were ordinary pronumerals I believed this question to be nonsense.

If a and b were pronumerals then I would have been correct BUT a and b are directional vectors so it is my answer that is nonsense. So at this point in time I will delete it. 

Thanks Tiggsy and Heureka and CPhill for the correction.  laugh

Melody  Oct 18, 2017
edited by Melody  Oct 19, 2017
edited by Melody  Oct 19, 2017
 #4
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+1

Sorry Melody, but that is wrong.

 

It's a perfectly acceptable exercise in vector addition.

a and b are vectors, they each have an magnitude, (unknown), and a direction. The direction of a is parallel to CA, in that direction and is parallel to and in the direction of CB.

Vectors obey a triangle law of addition.

In the original diagram, put a line between C and X, then the vector CX will equal the sum of the vectors CA and AX.

\(\displaystyle \underline{CX}=\underline{CA}+\underline{AX}\)

Similarly,

\(\displaystyle \underline{CX}=\underline{CB}+\underline{BX}\) ,

so long as you start and end at the same points, you can visit any other point(s) on the way.

The law can be extended to more than one intermediate point.

For example, put a line between Y and X, and you can say that \(\displaystyle \underline{BX}=\underline{BY}+\underline{YX}\),

so \(\displaystyle \underline{CX}=\underline{CB}+\underline{BY}+\underline{YX}\).

So long as you start and end at the same points, ... .

 

Onto the actual question.

We are told that \(\displaystyle \underline{AX}:\underline{XB}=1:2 \text{ , so, }2\underline{AX}=\underline{XB}\).

 

Using the triangle law,

\(\displaystyle \underline{CX}=\underline{CA}+\underline{AX}=3\underline{a}+\underline{AX}\),

and

\(\displaystyle \underline{CX}=\underline{CB}+\underline{BX}=\underline{CB}-\underline{XB}=6\underline{b}-2\underline{AX}\).

Eliminate the \(\displaystyle \underline{AX}\) between the last two equations, (twice the first plus the second), and we have

\(\displaystyle 3\underline{CX}=6\underline{a}+6\underline{b}\), so \(\displaystyle \underline{CX}=2(\underline{a}+\underline{b})\).

 

Using the triangle law again,

\(\displaystyle \underline{CY}=\underline{CB}+ \underline{BY}= 6\underline{b}+(5\underline{a}-\underline{b})=5(\underline{a}+\underline{b})\).

 

So, \(\displaystyle \underline{CX}= 2(\underline{a}+\underline{b})=\frac{2}{5}.5(\underline{a}+\underline{b})=\frac{2}{5}\text{ .}\underline{CY}\),

as required.

 

Tiggsy

Guest Oct 19, 2017
 #5
avatar+18715 
+1

please explain q.22

 

 

\(\text{CAYB is a quadrilateral. } \\ \vec{CA} = 3\vec{a} \\ \vec{CB} = 6\vec{b} \\ \vec{BY} = 5\vec{a}-\vec{b} \)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{\vec{CY}} {\vec{CX}}} \\\\ &=& \dfrac{\vec{BC}+\vec{YB}} {\vec{AC}+\vec{XA}} \\\\ &=& \dfrac{6\vec{b}+5\vec{a}-\vec{b}} {3\vec{a}+\frac13\vec{AB}} \\\\ &=& \dfrac{5\cdot(\vec{a}+\vec{b})} {3\vec{a}+\frac13(\vec{CB}-\vec{CA})} \\\\ &=& \dfrac{5\cdot(\vec{a}+\vec{b})} {3\vec{a}+\frac13(6\vec{b}-3\vec{a})} \\\\ &=& \dfrac{5\cdot(\vec{a}+\vec{b})} {3\vec{a}+2\vec{b}-\vec{a}} \\\\ &=& \dfrac{5\cdot(\vec{a}+\vec{b})} {2\cdot(\vec{a}+\vec{b})} \\\\ &=& \mathbf{\dfrac{5}{2}} \\ \hline \end{array}\)

 

laugh

heureka  Oct 19, 2017
 #6
avatar+91053 
+1

Yes I can now see I am wrong.

I took a and b to be pronumerals when they are cleaarly marked as directional vectors.

If a and b were ordinary pronumerals then the question would be nonsense but as directional vectors (as is stated in the question) it is my answer that is nonsense.  Sorry all.

 

Thanks Tiggsy and Heureka. 

 

 

Also CPhill found the answer as a video clip on a seperate site.

 

This is the site

http://www.onlinemathlearning.com/edexcel-gcse-specimen-p1-higher.html

Scroll down till you find this heading

"Edexcel GCSE 9 - 1 Specimen Paper 1 (No Calculator) Solutions for Questions 8 - 23"

Underneath it says "Show step by step solutions"

Click on that and the video will display.  This is question 22.  The answer for it begins at  55:27

You can forward the clip to bigin at that point. 

 

Thanks Chris for finding this resource for us :)

Melody  Oct 19, 2017

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