Question

If the range of the coordinate transformation f(x, y) = (-2x,-3y +1) is (4, -2), (2,-5),(-6,4), what

Answer 1

Answer:

The domain is ${(-2,1),(-1,2),(-3,\dfrac{5}{3})}$

Step-by-step explanation:

Given that,

The function is

$f(x,y)=(-2x,-3y+1)$

The coordinates of range are,

$range=(4,-2), (2,-5), (-6,4)}$

The domain is,

$Domain={(x_{1},y_{1}),(x_{2},y_{2}),(x_{3},y_{3})}$

We need to find the value of x₁ and y₁

Using given function

$f(x.y)=(-2x,-3y+1)$

$f(x_{1},y_{1})=(4,-2,)$

$(-2x_{1},-3y_{1}+1)=(4,-2)$

On equating value of x

$-2x_{1}=4$

$x_{1}=-2$

On equating value of y

$-3y_{1}+1=-2$

$-3y_{1}=-2-1$

$y_{1}=\dfrac{-2-1}{-3}$

$y_{1}=1$

We need to find the value of x₂ and y₂

Using given function

$f(x_{2},y_{2})=(2,-5,)$

$(-2x_{2},-3y_{2}+1)=(2,-5)$

On equating value of x

$-2x_{2}=2$

$x_{2}=-1$

On equating value of y

$-3y_{2}+1=-5$

$-3y_{2}=-5-1$

$y_{2}=\dfrac{-5-1}{-3}$

$y_{2}=2$

We need to find the value of x₃ and y₃

Using given function

$f(x_{3},y_{3})=(-6,4,)$

$(-2x_{3},-3y_{3}+1)=(-6,4)$

On equating value of x

$-2x_{3}=-6$

$x_{3}=-3$

On equating value of y

$-3y_{3}+1=-4$

$-3y_{3}=-4-1$

$y_{3}=\dfrac{-4-1}{-3}$

$y_{3}=\dfrac{5}{3}$

We need to find the domain

Using domain

$Domain={(-2,1),(-1,2),(-3,\dfrac{5}{3})}$

Hence, The domain is ${(-2,1),(-1,2),(-3,\dfrac{5}{3})}$