#### allysaleemally

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

#### Smile27

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 • Nivesh Ramlochun and Zsiddiqui

#### allysaleemally

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

#### Smile27

You're welcome #### allysaleemally

hey if the x axis is the invarient in a stretch, then will the stretch be parallel to the x axis?

#### Smile27

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 #### sagittarius94

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 :%)

#### arlery

Please discuss the exam 24 hours after the paper.

#### allysaleemally

yes paper 2 was extremely easy, paper 4 wiill be hard

#### haochen

yeah #### sagittarius94

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.

#### sagittarius94

and THANKYOU soooo much smile27.ur post related to shear and stretch is really helpful.thanks alot #### Aahmsil

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 !!! Cheers !!! :good: