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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.

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    $\begingroup$ Vertices are the intersections of the sides $\endgroup$ – lab bhattacharjee Mar 8 '16 at 15:08
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    $\begingroup$ I think your answers are on spot. $\endgroup$ – Win Vineeth Mar 8 '16 at 15:34
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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.

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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, $(-1,-3)$: $$\begin{cases}2x-3y-7=0 \\ 4x+y+7=0\end{cases}\quad \Rightarrow\quad x=-1,y=-3.$$ 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 $(5,1)$.

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