Maths IGCSE Help please

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Please im stuck on question 20b, maths paper 2, varient 3, october/nov 2010.

i dont understand the answer given on the marksheet.....please can anyone explain thiss to me.

thanks a million
 
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In the marking scheme, they give you the image that should be formed, i.e. the vertices of the triangle.

Let me tell you how to go about shearing the triangle :)

When it's x-axis invariant in a shear, it means the shear is parallel to the x-axis.

When it's y-axis invariant in a shear, it means the shear is parallel to y-axis.

Now that that's clear, let's proceed :)

Shear factor = [Distance Moved By the Point] divided by [Perpendicular Distance from the Invariant Line]

The shear factor does not change.
And, we have the perpendicular distance from the invariant line.

Now, we can find the distance moved by the point.

Point k (1 , 1)

Shear factor = [Distance Moved By the Point] divided by [Perpendicular Distance from the Invariant Line]
3 = x / 1
x = 3 .. So, the image of k will move 3 units to the left on the x-axis (it will move + 3 on the x-axis)

The co-ordinates for image of k ... (3+1 , 1) = (4 , 1)
(Notice, y doesn't change because shear is parallel to x ... Distance moved by the point is then added to the original co-ordinate of the point to get where to plot it...If you get what I mean..?)

Now do the same for the other 2 points.

Point M (1,3)
Shear factor = [Distance Moved By the Point] divided by [Perpendicular Distance from the Invariant Line]
3 = x / 3
x = 9 ... (9+1, 3) = (10,3)

Point L (5,1)
Shear factor = [Distance Moved By the Point] divided by [Perpendicular Distance from the Invariant Line]
3 = x / 1
x = 3 ... (3+5, 1) = (8,1)



Hope it's clear, if not, then just ask :)
 
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Smile27 said:
In the marking scheme, they give you the image that should be formed, i.e. the vertices of the triangle.

Let me tell you how to go about shearing the triangle :)

When it's x-axis invariant in a shear, it means the shear is parallel to the x-axis.

When it's y-axis invariant in a shear, it means the shear is parallel to y-axis.

Now that that's clear, let's proceed :)

Shear factor = [Distance Moved By the Point] divided by [Perpendicular Distance from the Invariant Line]

The shear factor does not change.
And, we have the perpendicular distance from the invariant line.

Now, we can find the distance moved by the point.

Point k (1 , 1)

Shear factor = [Distance Moved By the Point] divided by [Perpendicular Distance from the Invariant Line]
3 = x / 1
x = 3 .. So, the image of k will move 3 units to the left on the x-axis (it will move + 3 on the x-axis)

The co-ordinates for image of k ... (3+1 , 1) = (4 , 1)
(Notice, y doesn't change because shear is parallel to x ... Distance moved by the point is then added to the original co-ordinate of the point to get where to plot it...If you get what I mean..?)

Now do the same for the other 2 points.

Point M (1,3)
Shear factor = [Distance Moved By the Point] divided by [Perpendicular Distance from the Invariant Line]
3 = x / 3
x = 9 ... (9+1, 3) = (10,3)

Point L (5,1)
Shear factor = [Distance Moved By the Point] divided by [Perpendicular Distance from the Invariant Line]
3 = x / 1
x = 3 ... (3+5, 1) = (8,1)



Hope it's clear, if not, then just ask :)

That was really clear, understoood it
thankssssss alotttt
 
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I've attached a file, which has some basic points about each transformation. It helped make things clear for me, so I hope it does for you to.

Basically,

For Stretch.
y-axis is invariant --> stretch is parallel to x-axis.
x-axis is invariant --> stretch is parallel to y-axis.

For Shear,
y-axis is invariant --> shear is parallel to y-axis.
x-axis is invariant --> shear is parallel to x-axis.

Hope that helps :)
 
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I just gave my paper 2 maths may june 2011 today and wanted to tell that it was extremely easy;no questions on symmetry or transformation.there was a question on coordinate geometry,vectors,loci,inequalities,functons[obviously],matrix
i had some problem calculating upper bound and lower bound.i did solve it but m not sure if its correct.please help if anyone else attempted it today :%)
 
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paper 4......im having problems with a queston related to graphs of functions.its question 3,nov 2008[0580\o4]
thers no problem in plotting the graph but i cant solve part c [part 1].how can we find the three values of x where f(x)= -3x.i dont understand it.need help in the second part too.please if someone is available,do help me plzzz.
 
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My explanation

Well if it was a stretch, the triangle would have usually lengthened in one direction only - either parallel to the x-axis or parallel to the y-axis.

In case of a shear, as in the question, the object being sheared would appear to be made of different layers and each layer would have slid a different amount relative to a fixed line.

Simply speaking, if it was a stretch, the co-ordinates of triangle X would be (-1,2), (-4,2) and (-4,7).

However it is also important to accept the possibility that a stretch can take place in more than one direction but fortunately or unfortunately, I don't think that an IGCSE student is required to think to such an extent in that perspective as I've never seen a past paper question on transformations with an object being simultaneously stretched in two or more directions.

So, think simple !!! :)

Hope that helped !!! :D

Cheers !!! :good:
 
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