What is the set $f(D)$, i.e., the image of $D$ by $f$












1












$begingroup$


Let $1<a<2,0<b<1,c>1$ be real numbers. Let us consider the triangle $$D=(0,0)(c,0)(0,c-1)$$



and consider the function $f:ℝ²→ℝ²$ defined by



$$f(x,y)=(1-ax+y,bx)$$



My question is: What is the set $f(D)$, i.e., the image of $D$ by $f$.



I have no idea to start.






real-analysis geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried?
    $endgroup$
    – Test123
    Dec 29 '18 at 8:12










  • $begingroup$
    @Test123: I have no idea to start.
    $endgroup$
    – China
    Dec 29 '18 at 8:24










  • $begingroup$
    Try calculate the image of the three vertices of $D$ for the limiting values of your parameters. Then check where does $f$ sends lines to find the image of each side of $D$.
    $endgroup$
    – Test123
    Dec 29 '18 at 8:29










  • $begingroup$
    Can you convince yourself that the images of straight lines under this function will be straight lines? Then clearly the intersection of the straight lines will be preserved, which would imply that simply joining the images of the vertices is sufficient.
    $endgroup$
    – Uday Khanna
    Dec 29 '18 at 9:14










  • $begingroup$
    @UdayKhanna: Can you write this as a typical answer.
    $endgroup$
    – China
    Dec 29 '18 at 9:24
















1












$begingroup$


Let $1<a<2,0<b<1,c>1$ be real numbers. Let us consider the triangle $$D=(0,0)(c,0)(0,c-1)$$



and consider the function $f:ℝ²→ℝ²$ defined by



$$f(x,y)=(1-ax+y,bx)$$



My question is: What is the set $f(D)$, i.e., the image of $D$ by $f$.



I have no idea to start.






real-analysis geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried?
    $endgroup$
    – Test123
    Dec 29 '18 at 8:12










  • $begingroup$
    @Test123: I have no idea to start.
    $endgroup$
    – China
    Dec 29 '18 at 8:24










  • $begingroup$
    Try calculate the image of the three vertices of $D$ for the limiting values of your parameters. Then check where does $f$ sends lines to find the image of each side of $D$.
    $endgroup$
    – Test123
    Dec 29 '18 at 8:29










  • $begingroup$
    Can you convince yourself that the images of straight lines under this function will be straight lines? Then clearly the intersection of the straight lines will be preserved, which would imply that simply joining the images of the vertices is sufficient.
    $endgroup$
    – Uday Khanna
    Dec 29 '18 at 9:14










  • $begingroup$
    @UdayKhanna: Can you write this as a typical answer.
    $endgroup$
    – China
    Dec 29 '18 at 9:24














1












1








1


1



$begingroup$


Let $1<a<2,0<b<1,c>1$ be real numbers. Let us consider the triangle $$D=(0,0)(c,0)(0,c-1)$$



and consider the function $f:ℝ²→ℝ²$ defined by



$$f(x,y)=(1-ax+y,bx)$$



My question is: What is the set $f(D)$, i.e., the image of $D$ by $f$.



I have no idea to start.






real-analysis geometry







share|cite|improve this question











$endgroup$




Let $1<a<2,0<b<1,c>1$ be real numbers. Let us consider the triangle $$D=(0,0)(c,0)(0,c-1)$$



and consider the function $f:ℝ²→ℝ²$ defined by



$$f(x,y)=(1-ax+y,bx)$$



My question is: What is the set $f(D)$, i.e., the image of $D$ by $f$.



I have no idea to start.






real-analysis geometry




real-analysis geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 8:46







China

















asked Dec 29 '18 at 8:02









ChinaChina

1,4191029




1,4191029












  • $begingroup$
    What have you tried?
    $endgroup$
    – Test123
    Dec 29 '18 at 8:12










  • $begingroup$
    @Test123: I have no idea to start.
    $endgroup$
    – China
    Dec 29 '18 at 8:24










  • $begingroup$
    Try calculate the image of the three vertices of $D$ for the limiting values of your parameters. Then check where does $f$ sends lines to find the image of each side of $D$.
    $endgroup$
    – Test123
    Dec 29 '18 at 8:29










  • $begingroup$
    Can you convince yourself that the images of straight lines under this function will be straight lines? Then clearly the intersection of the straight lines will be preserved, which would imply that simply joining the images of the vertices is sufficient.
    $endgroup$
    – Uday Khanna
    Dec 29 '18 at 9:14










  • $begingroup$
    @UdayKhanna: Can you write this as a typical answer.
    $endgroup$
    – China
    Dec 29 '18 at 9:24


















  • $begingroup$
    What have you tried?
    $endgroup$
    – Test123
    Dec 29 '18 at 8:12










  • $begingroup$
    @Test123: I have no idea to start.
    $endgroup$
    – China
    Dec 29 '18 at 8:24










  • $begingroup$
    Try calculate the image of the three vertices of $D$ for the limiting values of your parameters. Then check where does $f$ sends lines to find the image of each side of $D$.
    $endgroup$
    – Test123
    Dec 29 '18 at 8:29










  • $begingroup$
    Can you convince yourself that the images of straight lines under this function will be straight lines? Then clearly the intersection of the straight lines will be preserved, which would imply that simply joining the images of the vertices is sufficient.
    $endgroup$
    – Uday Khanna
    Dec 29 '18 at 9:14










  • $begingroup$
    @UdayKhanna: Can you write this as a typical answer.
    $endgroup$
    – China
    Dec 29 '18 at 9:24
















$begingroup$
What have you tried?
$endgroup$
– Test123
Dec 29 '18 at 8:12




$begingroup$
What have you tried?
$endgroup$
– Test123
Dec 29 '18 at 8:12












$begingroup$
@Test123: I have no idea to start.
$endgroup$
– China
Dec 29 '18 at 8:24




$begingroup$
@Test123: I have no idea to start.
$endgroup$
– China
Dec 29 '18 at 8:24












$begingroup$
Try calculate the image of the three vertices of $D$ for the limiting values of your parameters. Then check where does $f$ sends lines to find the image of each side of $D$.
$endgroup$
– Test123
Dec 29 '18 at 8:29




$begingroup$
Try calculate the image of the three vertices of $D$ for the limiting values of your parameters. Then check where does $f$ sends lines to find the image of each side of $D$.
$endgroup$
– Test123
Dec 29 '18 at 8:29












$begingroup$
Can you convince yourself that the images of straight lines under this function will be straight lines? Then clearly the intersection of the straight lines will be preserved, which would imply that simply joining the images of the vertices is sufficient.
$endgroup$
– Uday Khanna
Dec 29 '18 at 9:14




$begingroup$
Can you convince yourself that the images of straight lines under this function will be straight lines? Then clearly the intersection of the straight lines will be preserved, which would imply that simply joining the images of the vertices is sufficient.
$endgroup$
– Uday Khanna
Dec 29 '18 at 9:14












$begingroup$
@UdayKhanna: Can you write this as a typical answer.
$endgroup$
– China
Dec 29 '18 at 9:24




$begingroup$
@UdayKhanna: Can you write this as a typical answer.
$endgroup$
– China
Dec 29 '18 at 9:24










1 Answer
1






active

oldest

votes


















3












$begingroup$

Okay, observe that by the linearity of the function it is evident that any straight line is being mapped to a straight line and thus it is sufficient to observe that the images of the vertices $(0,0),(c,0),(0,c-1)$, which are $(1,0),(1-ac,bc),(c,0)$ respectively determine the triangle that is f(D). Here there exist one common point $(c,0)$ between $D$ and the image of $D$.



Explicitly, If you break down the figure into lines and try to find the images of the lines then it is still a straightforward task. We have $D$ as the right angled triangle formed by the intersection of the lines which I will represent in their parametrised form



{$(t,0): 0<t<c$}, {$(t,(c-1)(1-frac{t}{c})):0<t<c$} , {$(0,t):0<t<c-1 $}



Now the images of these lines are



{$(1-at,bt): 0<t<c$}, {$(c-(a+frac{c-1}{c})t,bt):0<t<c$}, {$(1+t,0):0<t<c-1$}



And it can be easily observed that these three straight lines intersect at the three distinct points stated earlier forming the image triangle f(D).






share|cite|improve this answer











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    $begingroup$

    Okay, observe that by the linearity of the function it is evident that any straight line is being mapped to a straight line and thus it is sufficient to observe that the images of the vertices $(0,0),(c,0),(0,c-1)$, which are $(1,0),(1-ac,bc),(c,0)$ respectively determine the triangle that is f(D). Here there exist one common point $(c,0)$ between $D$ and the image of $D$.



    Explicitly, If you break down the figure into lines and try to find the images of the lines then it is still a straightforward task. We have $D$ as the right angled triangle formed by the intersection of the lines which I will represent in their parametrised form



    {$(t,0): 0<t<c$}, {$(t,(c-1)(1-frac{t}{c})):0<t<c$} , {$(0,t):0<t<c-1 $}



    Now the images of these lines are



    {$(1-at,bt): 0<t<c$}, {$(c-(a+frac{c-1}{c})t,bt):0<t<c$}, {$(1+t,0):0<t<c-1$}



    And it can be easily observed that these three straight lines intersect at the three distinct points stated earlier forming the image triangle f(D).






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      Okay, observe that by the linearity of the function it is evident that any straight line is being mapped to a straight line and thus it is sufficient to observe that the images of the vertices $(0,0),(c,0),(0,c-1)$, which are $(1,0),(1-ac,bc),(c,0)$ respectively determine the triangle that is f(D). Here there exist one common point $(c,0)$ between $D$ and the image of $D$.



      Explicitly, If you break down the figure into lines and try to find the images of the lines then it is still a straightforward task. We have $D$ as the right angled triangle formed by the intersection of the lines which I will represent in their parametrised form



      {$(t,0): 0<t<c$}, {$(t,(c-1)(1-frac{t}{c})):0<t<c$} , {$(0,t):0<t<c-1 $}



      Now the images of these lines are



      {$(1-at,bt): 0<t<c$}, {$(c-(a+frac{c-1}{c})t,bt):0<t<c$}, {$(1+t,0):0<t<c-1$}



      And it can be easily observed that these three straight lines intersect at the three distinct points stated earlier forming the image triangle f(D).






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        Okay, observe that by the linearity of the function it is evident that any straight line is being mapped to a straight line and thus it is sufficient to observe that the images of the vertices $(0,0),(c,0),(0,c-1)$, which are $(1,0),(1-ac,bc),(c,0)$ respectively determine the triangle that is f(D). Here there exist one common point $(c,0)$ between $D$ and the image of $D$.



        Explicitly, If you break down the figure into lines and try to find the images of the lines then it is still a straightforward task. We have $D$ as the right angled triangle formed by the intersection of the lines which I will represent in their parametrised form



        {$(t,0): 0<t<c$}, {$(t,(c-1)(1-frac{t}{c})):0<t<c$} , {$(0,t):0<t<c-1 $}



        Now the images of these lines are



        {$(1-at,bt): 0<t<c$}, {$(c-(a+frac{c-1}{c})t,bt):0<t<c$}, {$(1+t,0):0<t<c-1$}



        And it can be easily observed that these three straight lines intersect at the three distinct points stated earlier forming the image triangle f(D).






        share|cite|improve this answer











        $endgroup$



        Okay, observe that by the linearity of the function it is evident that any straight line is being mapped to a straight line and thus it is sufficient to observe that the images of the vertices $(0,0),(c,0),(0,c-1)$, which are $(1,0),(1-ac,bc),(c,0)$ respectively determine the triangle that is f(D). Here there exist one common point $(c,0)$ between $D$ and the image of $D$.



        Explicitly, If you break down the figure into lines and try to find the images of the lines then it is still a straightforward task. We have $D$ as the right angled triangle formed by the intersection of the lines which I will represent in their parametrised form



        {$(t,0): 0<t<c$}, {$(t,(c-1)(1-frac{t}{c})):0<t<c$} , {$(0,t):0<t<c-1 $}



        Now the images of these lines are



        {$(1-at,bt): 0<t<c$}, {$(c-(a+frac{c-1}{c})t,bt):0<t<c$}, {$(1+t,0):0<t<c-1$}



        And it can be easily observed that these three straight lines intersect at the three distinct points stated earlier forming the image triangle f(D).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 29 '18 at 10:13









        China

        1,4191029




        1,4191029










        answered Dec 29 '18 at 9:58









        Uday KhannaUday Khanna

        21718




        21718






























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