the equation of two sides of a parallelogram are $2x-3y+7=0$ and $4x+y-21=0$ and one vertex is $(-1,-3)$....












2












$begingroup$


First, I checked if the point $(-1,-3)$ is not a solution to the two given equations above so therefore none of those lines passes that point. Then, I solved for the lines parallel to equations above getting $2x-3y-7=0$ and $4x+y+7=0$ respectively.



Edit: i managed to find the other vertices which are (5,1), (4,5), and (-2,1). I just want to ask if my vertices are correct.






analytic-geometry







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  • 1




    $begingroup$
    Vertices are the intersections of the sides
    $endgroup$
    – lab bhattacharjee
    Mar 8 '16 at 15:08






  • 1




    $begingroup$
    I think your answers are on spot.
    $endgroup$
    – Win Vineeth
    Mar 8 '16 at 15:34
















2












$begingroup$


First, I checked if the point $(-1,-3)$ is not a solution to the two given equations above so therefore none of those lines passes that point. Then, I solved for the lines parallel to equations above getting $2x-3y-7=0$ and $4x+y+7=0$ respectively.



Edit: i managed to find the other vertices which are (5,1), (4,5), and (-2,1). I just want to ask if my vertices are correct.






analytic-geometry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Vertices are the intersections of the sides
    $endgroup$
    – lab bhattacharjee
    Mar 8 '16 at 15:08






  • 1




    $begingroup$
    I think your answers are on spot.
    $endgroup$
    – Win Vineeth
    Mar 8 '16 at 15:34














2












2








2





$begingroup$


First, I checked if the point $(-1,-3)$ is not a solution to the two given equations above so therefore none of those lines passes that point. Then, I solved for the lines parallel to equations above getting $2x-3y-7=0$ and $4x+y+7=0$ respectively.



Edit: i managed to find the other vertices which are (5,1), (4,5), and (-2,1). I just want to ask if my vertices are correct.






analytic-geometry







share|cite|improve this question











$endgroup$




First, I checked if the point $(-1,-3)$ is not a solution to the two given equations above so therefore none of those lines passes that point. Then, I solved for the lines parallel to equations above getting $2x-3y-7=0$ and $4x+y+7=0$ respectively.



Edit: i managed to find the other vertices which are (5,1), (4,5), and (-2,1). I just want to ask if my vertices are correct.






analytic-geometry




analytic-geometry






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share|cite|improve this question













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share|cite|improve this question








edited Mar 8 '16 at 15:30







enggfrosh

















asked Mar 8 '16 at 15:06









enggfroshenggfrosh

115




115








  • 1




    $begingroup$
    Vertices are the intersections of the sides
    $endgroup$
    – lab bhattacharjee
    Mar 8 '16 at 15:08






  • 1




    $begingroup$
    I think your answers are on spot.
    $endgroup$
    – Win Vineeth
    Mar 8 '16 at 15:34














  • 1




    $begingroup$
    Vertices are the intersections of the sides
    $endgroup$
    – lab bhattacharjee
    Mar 8 '16 at 15:08






  • 1




    $begingroup$
    I think your answers are on spot.
    $endgroup$
    – Win Vineeth
    Mar 8 '16 at 15:34








1




1




$begingroup$
Vertices are the intersections of the sides
$endgroup$
– lab bhattacharjee
Mar 8 '16 at 15:08




$begingroup$
Vertices are the intersections of the sides
$endgroup$
– lab bhattacharjee
Mar 8 '16 at 15:08




1




1




$begingroup$
I think your answers are on spot.
$endgroup$
– Win Vineeth
Mar 8 '16 at 15:34




$begingroup$
I think your answers are on spot.
$endgroup$
– Win Vineeth
Mar 8 '16 at 15:34










2 Answers
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$begingroup$

The parallelogram has $4$ vertices. The vertex $(-1,-3)$ is not an element of both given equations. The $2$ equations given are not parallel to eachother.



This tells us that the equations are from 2 abutting sides. These equations can only obtain $3$ vertices, so the given vertex is the $4^{th}$ vertex. You've already solved it almost. You calculated the equations of the other $2$ sides.



By calculating the $3$ remaining intersection points of the non-parallel equations you've found your other vertices.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You just need to intersect each of the sides to obtain the vertices. For instance, from $2x-3y-7=0$ and $4x+y+7=0$ you have the point that you already know, $(-1,-3)$:
    $$begin{cases}2x-3y-7=0 \ 4x+y+7=0end{cases}quad Rightarrowquad x=-1,y=-3.$$
    Try with the other intersecting pairs of lines:




    • Second vertex: $2x-3y-7=0$ and $4x+y-21=0$

    • Third vertex: $2x-3y+7=0$ and $4x+y-21=0$

    • Fourth vertex: $2x-3y+7=0$ and $4x+y+7=0$


    Solution:




    The vertices are $(-1,-3)$, $(-2,1)$, $(4,5)$ and $(5,1)$.







    share|cite|improve this answer











    $endgroup$













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      2 Answers
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      2 Answers
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      0












      $begingroup$

      The parallelogram has $4$ vertices. The vertex $(-1,-3)$ is not an element of both given equations. The $2$ equations given are not parallel to eachother.



      This tells us that the equations are from 2 abutting sides. These equations can only obtain $3$ vertices, so the given vertex is the $4^{th}$ vertex. You've already solved it almost. You calculated the equations of the other $2$ sides.



      By calculating the $3$ remaining intersection points of the non-parallel equations you've found your other vertices.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        The parallelogram has $4$ vertices. The vertex $(-1,-3)$ is not an element of both given equations. The $2$ equations given are not parallel to eachother.



        This tells us that the equations are from 2 abutting sides. These equations can only obtain $3$ vertices, so the given vertex is the $4^{th}$ vertex. You've already solved it almost. You calculated the equations of the other $2$ sides.



        By calculating the $3$ remaining intersection points of the non-parallel equations you've found your other vertices.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          The parallelogram has $4$ vertices. The vertex $(-1,-3)$ is not an element of both given equations. The $2$ equations given are not parallel to eachother.



          This tells us that the equations are from 2 abutting sides. These equations can only obtain $3$ vertices, so the given vertex is the $4^{th}$ vertex. You've already solved it almost. You calculated the equations of the other $2$ sides.



          By calculating the $3$ remaining intersection points of the non-parallel equations you've found your other vertices.






          share|cite|improve this answer









          $endgroup$



          The parallelogram has $4$ vertices. The vertex $(-1,-3)$ is not an element of both given equations. The $2$ equations given are not parallel to eachother.



          This tells us that the equations are from 2 abutting sides. These equations can only obtain $3$ vertices, so the given vertex is the $4^{th}$ vertex. You've already solved it almost. You calculated the equations of the other $2$ sides.



          By calculating the $3$ remaining intersection points of the non-parallel equations you've found your other vertices.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 8 '16 at 15:28









          RutsRuts

          575319




          575319























              0












              $begingroup$

              You just need to intersect each of the sides to obtain the vertices. For instance, from $2x-3y-7=0$ and $4x+y+7=0$ you have the point that you already know, $(-1,-3)$:
              $$begin{cases}2x-3y-7=0 \ 4x+y+7=0end{cases}quad Rightarrowquad x=-1,y=-3.$$
              Try with the other intersecting pairs of lines:




              • Second vertex: $2x-3y-7=0$ and $4x+y-21=0$

              • Third vertex: $2x-3y+7=0$ and $4x+y-21=0$

              • Fourth vertex: $2x-3y+7=0$ and $4x+y+7=0$


              Solution:




              The vertices are $(-1,-3)$, $(-2,1)$, $(4,5)$ and $(5,1)$.







              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                You just need to intersect each of the sides to obtain the vertices. For instance, from $2x-3y-7=0$ and $4x+y+7=0$ you have the point that you already know, $(-1,-3)$:
                $$begin{cases}2x-3y-7=0 \ 4x+y+7=0end{cases}quad Rightarrowquad x=-1,y=-3.$$
                Try with the other intersecting pairs of lines:




                • Second vertex: $2x-3y-7=0$ and $4x+y-21=0$

                • Third vertex: $2x-3y+7=0$ and $4x+y-21=0$

                • Fourth vertex: $2x-3y+7=0$ and $4x+y+7=0$


                Solution:




                The vertices are $(-1,-3)$, $(-2,1)$, $(4,5)$ and $(5,1)$.







                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You just need to intersect each of the sides to obtain the vertices. For instance, from $2x-3y-7=0$ and $4x+y+7=0$ you have the point that you already know, $(-1,-3)$:
                  $$begin{cases}2x-3y-7=0 \ 4x+y+7=0end{cases}quad Rightarrowquad x=-1,y=-3.$$
                  Try with the other intersecting pairs of lines:




                  • Second vertex: $2x-3y-7=0$ and $4x+y-21=0$

                  • Third vertex: $2x-3y+7=0$ and $4x+y-21=0$

                  • Fourth vertex: $2x-3y+7=0$ and $4x+y+7=0$


                  Solution:




                  The vertices are $(-1,-3)$, $(-2,1)$, $(4,5)$ and $(5,1)$.







                  share|cite|improve this answer











                  $endgroup$



                  You just need to intersect each of the sides to obtain the vertices. For instance, from $2x-3y-7=0$ and $4x+y+7=0$ you have the point that you already know, $(-1,-3)$:
                  $$begin{cases}2x-3y-7=0 \ 4x+y+7=0end{cases}quad Rightarrowquad x=-1,y=-3.$$
                  Try with the other intersecting pairs of lines:




                  • Second vertex: $2x-3y-7=0$ and $4x+y-21=0$

                  • Third vertex: $2x-3y+7=0$ and $4x+y-21=0$

                  • Fourth vertex: $2x-3y+7=0$ and $4x+y+7=0$


                  Solution:




                  The vertices are $(-1,-3)$, $(-2,1)$, $(4,5)$ and $(5,1)$.








                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Mar 8 '16 at 15:39

























                  answered Mar 8 '16 at 15:31









                  AugSBAugSB

                  3,39921734




                  3,39921734






























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