Q42. - Algebra 2

Question

Find the coordinates of the vertices of the parallelogram whose sides are contained in the lines whose equations are $2x+y=-12$, $2x-y=-8$,$2x-y-4=0$ and $4x+2y=24$.

Step-by-Step Solution

Answer

The vertices of the parallelogram are A$\left(-2 , -8\right)$, B$\left(-5,-2\right)$, C$\left(4,4\right)$, and D$\left(1,10\right)$.

Step-1 – Apply the concept of parallelogram

The two-dimensional geometrical shape whose opposite sides are parallel to each other and equal in length is called a parallelogram. The interior opposite angles is also equal of a parallelogram.

To find the vertices through the equations, solve the equations first and then plot them on graph. The intersecting point between the lines will be the vertices of the parallelogram.

Step-2 – Solving the equations

Consider the first equation.

$2x+y=-12$

To solve, first substitute x as zero to find the value of y and then substitute y as zero to find the value of x as shown. 

When $x=0$,

$\begin{array}{c}2x+y=-12\\ 2\left(0\right)+y=-12\\ 0+y=-12\\ y=-12\end{array}$

When $y=0$,

$\begin{array}{c}2x+y=-12\\ 2x+0=-12\\ 2x=-12\\ x=-6\end{array}$

$\begin{array}{ccc}x& 0& -6\\ y& -12& 0\end{array}$

Repeat the same steps to solve all other equations.

Take second equation now.

$2x-y=-8$

To solve, first substitute x as zero to find the value of y and then substitute y as zero to find the value of x as shown. 

When $x=0$,

$\begin{array}{c}2x-y=-8\\ 2\left(0\right)-y=-8\\ 0-y=-8\\ y=8\end{array}$

When $y=0$,

$\begin{array}{c}2x-y=-8\\ 2x-0=-8\\ 2x=-8\\ x=-4\end{array}$

The solution of the second equation is:

$\begin{array}{ccc}x& 0& -4\\ y& 8& 0\end{array}$

Let us take third equation now.

$2x-y-4=0$

To solve, first substitute x as zero to find the value of y and then substitute y as zero to find the value of x as shown. 

When $x=0$,

$\begin{array}{c}2x-y-4=0\\ 2\left(0\right)-y-4=0\\ 0-y=4\\ y=-4\end{array}$
When $y=0$,

$\begin{array}{c}2x-y-4=0\\ 2x-0-4=0\\ 2x=4\\ x=2\end{array}$

The solution of the third equation is:

$\begin{array}{ccc}x& 0& 2\\ y& -4& 0\end{array}$

Now solve the fourth equation.

$4x+2y=24$

To solve, first substitute x as zero to find the value of y and then substitute y as zero to find the value of x as shown. 

When $x=0$,

$\begin{array}{c}4x+2y=24\\ 4\left(0\right)+2y=24\\ 0+2y=24\\ y=12\end{array}$

When $y=0$,

$\begin{array}{c}4x+2y=24\\ 4x+2\left(0\right)=24\\ 4x=24\\ x=6\end{array}$

The solution of the fourth equation is:

$\begin{array}{ccc}x& 0& 6\\ y& 12& 0\end{array}$

Step-3 – Graphing the coordinates

The solutions of all the equations we got are:

Now plot all the coordinates on the graph and obtain the coordinates of the points where these lines intersect with each other.

The red line represents the equation $2x+y=-12$. The blue line represents the equation $2x-y=-8$. The green line represents the equation $2x-y-4=0$. And the purple line represents the equation $4x+2y=24$.

The points where these lines meet are the vertices of the parallelogram that is A$\left(-2 , -8\right)$, B$\left(-5,-2\right)$, C$\left(4,4\right)$, and D$\left(1,10\right)$ represents the vertices of the parallelogram.

 

Hence, the vertices of the parallelogram are A$\left(-2 , -8\right)$, B$\left(-5,-2\right)$, C$\left(4,4\right)$, and D$\left(1,10\right)$.