Derive the equation of a line after x+(1/2)y =0 is mapped by T

[Richie Smash]

Homework Statement

Hello hello, I am on the second to last question of 6 practice examinations I've been doing!

''There is a transformation matrix T =
\begin{pmatrix}
4 & 9\\
3& 4
\end{pmatrix}

They want me to derive the equation of the line onto which the line x +(1/2)y = 0 is mapped by the transformation T.

Homework Equations

From x+1/2y=0
y= -2x

The Attempt at a Solution

So first I rearranged the equation to solve for (Y) and got
y= -2x

So then I multiplied each element in the matrix by - 2x and got
\begin{pmatrix}
-8x & -18x\\
-6x& -8x
\end{pmatrix}

But this is confusing, they want the equation of the line, to be honest I don't know which direction to go here, do I now multiply this new Matrix by the original T?

I tried that and got
\begin{matrix}
-86x &-144x \\
-48x& -86x
\end{matrix}

I'm not sure which direction to take, as they want the equation of the line...

 

[FactChecker]

After the transformation to new coordinates (x', y'), you have an equation for each of x' and y' in terms of x. You should be able to write those two equations down and solve the simultaneous equations for y' in terms of x'.

 

[Richie Smash]

Sorry, I'm not quite understanding what you're saying?
:(

 

[Ray Vickson]

Sorry, I'm not quite understanding what you're saying?
:(

He is saying that the transformation gives
$$\begin{array}{l}
x' = 4x + 3y \\
y' = 9x+4y
\end{array}
$$
If you know that $y = -2x$ you can figure out how $x'$ and $y'$ are related.

 

[Mark44]

So first I rearranged the equation to solve for (Y) and got
y= -2x

This line is generated by all constant multiples of the vector <1, -2>. The parametric form of this line would be <t, -2t>, for real numbers t. Your transformation matrix operates on vectors, and produces vectors as its outputs.

 

[Richie Smash]

I'm not sure I understand this method, but here is something that I follow:

I take the knowledge that when x = 1 , y= -2, and when x=2, y= -4

I put these under the transformation and obtain the two images of the points (-14, -9) and (-28,-10)

I then find the gradient which is (1/14)x and substitute the gradient and one of the new image points into the equation y=mx +c

So I'd get c = -8

SO the final equation of the new line would be y=(1/14)x-8

 

[FactChecker]

SO the final equation of the new line would be y=(1/14)x-8

You should check this answer back in your original equations. Since y = -2x, you know that the original line goes through x=0, y=0. The transformation will map that to x'=0, y'=0. Your answer for the new line does not go through (0,0), so it must be wrong.

Your general method is ok. The second point you used x=2, y=-2 is not correct. It should be x=2, y=-4.

 

[Richie Smash]

Ah yes... you'r right, I am defeated by this question I'm afraid, I just don't knwo where to begin?

 

[FactChecker]

Ah yes... you'r right, I am defeated by this question I'm afraid, I just don't knwo where to begin?

You're very close to getting the answer. Your general method is ok. The second point you used x=2, y=-2 is not correct. It should have been x=2, y=-4. Or you can use x=0, y=0, which may be simpler.

 

[Richie Smash]

SOLVED :D

 

[SammyS]

SOLVED :D

Before marking this thread as Solved, it would be helpful for you to give the solution.

Somebody may come along later and be interested in this question and its solution.

 

[Richie Smash]

Ok well the solution was, because the original line goes through zero, I can use the two points (0,0) and (1,-2)

From here I ran them through the matrix and obtained the image points (0,0) (-14,-5)
I found the gradient for these points and then I used either point to sub into the equation of a straight line y=mx +c using my newly found gradient as well.

From here I calculated the equation of the line to be y=(5/14)x

 

[FactChecker]

You should check the (-14, -9) point again. That looks wrong. I agree with your final answer, so I think that you just forgot to record a correction of that point.

 

[Richie Smash]

fixed

 

[Ray Vickson]

Ok well the solution was, because the original line goes through zero, I can use the two points (0,0) and (1,-2)

From here I ran them through the matrix and obtained the image points (0,0) (-14,5)
I found the gradient for these points and then I used either point to sub into the equation of a straight line y=mx +c using my newly found gradient as well.

From here I calculated the equation of the line to be y=(5/14)x

Answer is OK, but you really should not use the same symbols (x,y) in both cases. The transformation gives a change of coordinates, from (x,y) to something else, such as (x', y'). You want the line in the NEW coordinate system, so it is y' = (5/14)x'. Never, use the exact same symbol for two different quantities in the same problem.

 

[FactChecker]

fixed

Nope. Still a problem. It's important to be careful -- almost machine-like on these calculations. You can see that if there are a long string of calculations, the odds of getting it correct all the way through depends on being very careful at each step.
UPDATE: Good. Post #12 now looks correct to me.
PS. The need to be so careful is why I turned to computer programming -- once you get the code right, the calculations are correct every time no matter how complicated and tricky they are.