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I already saw this question here, but the answers are not clear for me. Hope I could get clear solution and answer. I already got the $2^{nd}$ vertex which is $(4, 5)$. How do I get the other two vertices? Thank you!

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One approach is to rewrite the line $2x - 3y = -7$ as

$$L_1 = \begin{bmatrix} 4 \\ 5 \end{bmatrix} + \begin{bmatrix} 3 \\ 2 \end{bmatrix}t$$

and rewrite $4x + y = 21$ as

$$L_2 = \begin{bmatrix} 4 \\ 5 \end{bmatrix} + \begin{bmatrix} 1 \\ -4 \end{bmatrix}s$$

Then solve

$$\begin{bmatrix} 4 \\ 5 \end{bmatrix} + \begin{bmatrix} 3 \\ 2 \end{bmatrix}t + \begin{bmatrix} 1 \\ -4 \end{bmatrix}s = \begin{bmatrix} -1 \\ 3 \end{bmatrix}$$

to get $t = -2, s = 1$ making the other 2 points:

$$P_1 = \begin{bmatrix} 4 \\ 5 \end{bmatrix} + \begin{bmatrix} 3 \\ 2 \end{bmatrix}\cdot -2 = \begin{bmatrix} -2 \\ 1 \end{bmatrix}$$

$$P_2 = \begin{bmatrix} 4 \\ 5 \end{bmatrix} + \begin{bmatrix} 1 \\ -4 \end{bmatrix} \cdot 1 = \begin{bmatrix} 5 \\ 1 \end{bmatrix}$$

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By your equations, you can see that $(-1,-3)$ isn't a point of neither one of your two sides (lines), and $(4,5)$ is a point of both, so you need to find two more sides, one parallel to the first line that pass through $(-1,-3)$ and anohter parallel to the second that also pass through $(-1,-3)$, this image helps:enter image description here

Here $f: 2x-3y+7=0$. Now, lets find these lines, the first one, it must be parallel to $2x-3y+7=0$ (that's the same of $y=\frac{2}{3}x + \frac{7}{3}$), to be parallel it must have the same linear coefficient, so it's of the form $y=\frac{2}{3}x+b$ since it pass through $(-1,-3)$: $$-3= -\frac{2}{3}+b \Rightarrow b=-\frac{7}{3} $$ So you have found onde of the sides $y=\frac{2}{3}x -\frac{7}{3}$ ( $3y-2x+7=0$). To find the other side you do the same thing, but now you find a line that's parallel to $4x+y-21=0$. When you find your sides, their intersectons with the sides you already have are the vetices you are looking for.

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HINT:

Let $AB:2x-3y+7=0$ and $AD:4x+y=21$

Can you find $A?$

Clearly, $B(-1,-3)$

Now as $AD||BC$ and $BC$ passes through $B, BC:4x+y=-15$

So, the coordinates of $D,C$ will be $(a,21-4a);(b,-15-4b)$ respectively.

Finally use "In a Parallelogram, the Diagonals Bisect Each Other" i.e., the midpoint of $BD, AC$ will be same.

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The vertex $A(-1,-3)$ isn’t on either of the given lines, so it and the other vertex that you’ve found, $C(4,5)$ are the ends of one of the parallelogram’s diagonals. There are a few ways to find the remaining two vertices.

One way is to find the equation of one of the missing sides and compute its intersection $B$ with the side to which it’s not parallel. The fourth vertex $D$ can then be found by using the fact that $A-B=D-C$.

You can determine the equation of a line parallel to a given one almost mechanically. The equation $ax+by+c=0$ can be written in vector form as $\mathbf n\cdot\mathbf x=const.$, where $\mathbf n=(a,b)$ is a vector that’s perpendicular (normal) to the line. Since this equation holds for all points on the line, the constant on the right-hand side is equal to $\mathbf n\cdot\mathbf p$, where $\mathbf p$ is a known point on the line. Applying this to the line $2x-3y+7=0$, we find that the parallel line through $(-1,-3)$ is $(2,-3)\cdot(x,y)=(2,-3)\cdot(-1,-3)$, which, after multiplying things out and rearranging, is $2x-3y-7=0$. With this equation in hand, you can find the two missing vertices as described above.

Another way is to use coordinate transformations. Suppose that a vertex of the parallelogram were at the origin and its sides parallel to the coordinate axes. Then the problem is easy: it’s a rectangle with vertices at $(x',0)$, $(0,y')$ and $(x',y')$. So, we switch to a coordinate system in which the parallelogram we’re building looks like a rectangle, find these points in that coordinate system, and switch back. First, translate so that the known point $A(-1,-3)$ becomes the new origin. Direction vectors for the two given lines are $\mathbf v_1=(-3,-2)$ and $\mathbf v_2=(-1,4)$, respectively, which become the columns of the matrix $M$ for the transformation to the current coordinate system. Transform $C-A$ by the inverse of this matrix: $$M^{-1}(C-A)=\frac1{14}\pmatrix{4&1\\2&-3}\pmatrix{5\\8}=\pmatrix{2\\1}.$$ The coordinates of the two missing vertices in the original frame are thus $(2,0)$ and $(0,1)$ multiplied by $M$, that is, $A+2\mathbf v_1=(5,1)$ and $A+1\mathbf v_2=(-2,1)$.

Note that with this second method, you don’t even need to find point $C$ first. Since the two known lines will be parallel to the new coordinate axes, you can use any points on them to get the necessary $x'$ and $y'$ values. By inspection, we can see that $(7,7)$ is on the first line and $(7,-7)$ is on the second. Subtracting $A$ from these points and transforming them by $M^{-1}$: $$\frac1{14}\pmatrix{4&1\\-2&3}\pmatrix{8&8\\10&-4}=\pmatrix{\cdots&2\\1&\cdots}.$$ We only need the $y'$ coordinate of the first point and the $x'$ coordinate of the second, so I haven’t bothered to compute the other entries in this product. Just as in the previous paragraph, we then have $A+2\mathbf v_1$ and $A+1\mathbf v_2$ for the two vertices adjacent to $A$, and $A+2\mathbf v_1+1\mathbf v_2$ for the opposite vertex, which you can verify is $(4,5)$, just as you had already calculated.

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