#1

*f*: (x,y) -> (x',y'), where x' = 2x-y and y' = x-y is a transformation.

L and K are lines with equations x - 3y - 5=0 and 3x + y + 1=0.

i) Prove that L is perpendicular to K

ii) Find the equations of

*f*(L) and

*f*(K).

iii) Investigate if

*f*(L) is perpendicular to

*f*(K).

____

Ok, we started this crap today, but I missed half the class, and I've NO idea what to do.... help!?

#2

how come you have classes already ?

I knew that , but it was like 2 years ago .. :/

I knew that , but it was like 2 years ago .. :/

#3

how come you have classes already ?

I knew that , but it was like 2 years ago .. :/

how come you dont, classes have been on for like 2-3 weeks, depending on the school

the first one, you have to show that their slopes are negative inverses

like if one has a slope of 2, and the other has a slope of -1/2

#4

Alright:

1. Look at the math book

2. Ask someone who paid attention.

3.

1. Look at the math book

2. Ask someone who paid attention.

3.

*NEVER*ask the Pit a math question.
#5

Well, you could always start by isolating y in both equations, and trying to get it in the form of

y = ax + b

If you then multiply the a's of both equations, they're perpendicular if the result is -1.

EDIT: Shit, I wasn't supposed to help, was I?

y = ax + b

If you then multiply the a's of both equations, they're perpendicular if the result is -1.

EDIT: Shit, I wasn't supposed to help, was I?

#6

the anser is false

#7

Alright:

1. Look at the math book

2. Ask someone who paid attention.

3.NEVERask the Pit a math question.

The pit is full of maths geeks.

Myself included

(rearrange the equations of the lines into the form y=mx+c, if the two ms multiply to make -1 then theyre perpendicular.)

#8

Yeah I got that. The perpendicular shizzle is easy, it's the transformed equations that are difficult

*Last edited by blackenedktulu at Sep 3, 2008,*

#9

Sub things in, simple as.

Part 1: Use M1M2=-1 having rearranged L and K into the form Y=MX+C, or having differentiated if you're into the whole parametric differentiation thing. Or isolate y and differentiate normally.

Part 2: Use the transformation instructions in the question to transform each equation.

As in "f: (x,y) -> (x',y'), where x' = 2x-y and y' = x-y is a transformation."

For L: x - 3y - 5=0

x goes to x', y goes to y' since the f: part says so.

You know what these are, so sub them in:

f(L)= (2x-y)-3(x-y)-5=0.

Then simplify, and repeat for K.

Part 3's the same as part 1, but using f(L) and f(K) as found in part 2. If this goes wrong, check your part 2.

Part 1: Use M1M2=-1 having rearranged L and K into the form Y=MX+C, or having differentiated if you're into the whole parametric differentiation thing. Or isolate y and differentiate normally.

Part 2: Use the transformation instructions in the question to transform each equation.

As in "f: (x,y) -> (x',y'), where x' = 2x-y and y' = x-y is a transformation."

For L: x - 3y - 5=0

x goes to x', y goes to y' since the f: part says so.

You know what these are, so sub them in:

f(L)= (2x-y)-3(x-y)-5=0.

Then simplify, and repeat for K.

Part 3's the same as part 1, but using f(L) and f(K) as found in part 2. If this goes wrong, check your part 2.

*Last edited by MopMaster at Sep 3, 2008,*