# Linear Fractional Transformation - find the formula

http://math.sfsu.edu/federico/Clase/Math350.S15/linea.JPG [Broken] 1. Homework Statement
The picture below represents the map from a "green projective line" to a "red projective line." It takes the "green points" 1,3,7,-11 to the "red points" 0,6,10,20, respectively as shown by the ruler. Let f be the corresponding linear fractional, so f(1) = 0, f(3) = 6, f(7) = 10, f(-11) = 20. Find a formula for the "red point" f(x) on the ruler where the "green point" x lands.

## Homework Equations

a linear fractional is where f(x) = (ax+b)/(cx+d) where ad-bc is not equal to 0.

## The Attempt at a Solution

First a did a system of equations:
(a(1) + b) / ( c(1) + d ) = 0
(a(3) +b) / ( c(3) + d )= 6
(a(7) +b) / ( c(7) + d ) = 10
(a(-11) +b) / (c(-11) + d )= 20

Thinking I could solve for a,b,c,d. However if I get all the letters in terms of let's say d and solve, this becomes nonsensical, since something d over something d is a number (the d's reduce) , and I'm left with a number = another number.

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Svein
a linear fractional is where f(x) = (ax+b)/(cx+d) where ad-bc is not equal to 0.
Since you have a fraction, you can choose either a or c to be 1. So, choose c = 1. The first equation says a⋅1 + b = 0, so b = -a. That's two variables accounted for. Now the rest is easy...

Choosing a or c to be 1 (or any number) seems to be an invalid assumption since even this is a ratio of a sum not just of a number over a number (choosing c=1 affects d, and choosing a=1 affects b), but please explain if I am incorrect. I chose c to be 1, then since I had everything else worked out in terms of d I plugged it all in. When I did, c=1 did not give f(3) = 6, so I solved for what c would be and got c = 1/6. Then, I tried to test this formula with f(7) = 10. It didn't work.

Svein
Choosing a or c to be 1 (or any number) seems to be an invalid assumption since even this is a ratio of a sum not just of a number over a number (choosing c=1 affects d, and choosing a=1 affects b), but please explain if I am incorrect. I chose c to be 1, then since I had everything else worked out in terms of d I plugged it all in. When I did, c=1 did not give f(3) = 6, so I solved for what c would be and got c = 1/6. Then, I tried to test this formula with f(7) = 10. It didn't work.
I am sorry, but you have made an error somewhere. One solution is: a = 15, b= -15, c=1, d=2.

About choosing one of the variables: A fundamental fact of fractions is that you can multiply or divide by the same number above or below. Therefore $\frac{a\cdot x + b}{c\cdot x + d}=\frac{a(x+\frac{b}{a})}{a(\frac{c\cdot x}{a}+\frac{d}{a})}=\frac{c(\frac{a}{c}x+\frac{b}{c})}{c(x+\frac{d}{c})}$ (of course, you cannot do that if the one you choose turns out to be 0).

Thanks very much Svein.

Mentallic
Homework Helper
Another quick simplification is to notice the case where f(x)=0

$$\frac{a+b}{c+d}=0$$
and for a fraction to equal 0, it means that the numerator = 0, hence

$$a+b=0$$

$$a=-b\text{, or } b=-a$$

Therefore you just have an equation of the form

$$f(x)=\frac{ax-a}{cx+d}$$