Solve graphically the system of equations 2x – 5y + 4 = 0 2x + y - 8 = 0. Find the coordinates of the vertices of the triangle formed by these two lines and the y-axis.

Home » Solve graphically the system of equations 2x – 5y + 4 = 0 2x + y – 8 = 0. Find the coordinates of the vertices of the triangle formed by these two lines and the y-axis.

Solve graphically the system of equations
2x – 5y + 4 = 0
2x + y – 8 = 0. Find the coordinates of the vertices of the triangle formed by these two lines and the y-axis.

Solve graphically the system of equations
2x – 5y + 4 = 0
2x + y – 8 = 0. Find the coordinates of the vertices of the triangle formed by these two lines and the y-axis.

Solution:

Draw a horizontal line on a graph paper $X^{\prime} O X$ and a vertical line YOY’ as the $x$ -axis and $y$ -axis, respectively.
$\begin{array}{l} \quad \text { Graph of } 2 x-5 y+4=0 \\ 2 x-5 y+4=0 \\ \Rightarrow 5 y=(2 x+4) \\ \therefore y=\frac{2 x+4}{5} \\ \ldots \ldots \ldots . . \end{array}$

Putting $x=3$ , we obtain $y=2$
Putting $x=-2$ , we obtain $y=0$ .
Putting $x=8$ , we obtain $y=4$ .

Therefore, we have the following table for the equation $2 x-5 y+4=0$

$\begin{tabular}{|c|c|c|c|} \hline$x$ & $-2$ & 3 & 8 \\ \hline$y$ & 0 & 2 & 4 \\ \hline \end{tabular}$

Now, plot the points $A(-2,0), B(3,2)$ and $C(8,4)$ on the graph paper.
Join $A B$ and $B C$ to get the graph line $A C$ . Extend it on both ways.

So, $A C$ is the graph of $2 x-5 y+4=0$ .
$\begin{array}{l} \text { Graph of } 2 x+y-8=0 \\ 2 x+y-8=0 \\ \Rightarrow y=(-2 x+8) \ldots \ldots \ldots . . \end{array}$

Putting $x=3$ , we obtain $y=2$ .
Putting $x=1$ , we obtain $y=6$ .
Putting $x=2$ , we obtain $y=4$ .

Therefore, we have the following table for the equation $2 x+y-8=0$

$\begin{tabular}{|c|c|c|c|} \hline$x$ & 1 & 3 & 2 \\ \hline$y$ & 6 & 2 & 4 \\ \hline \end{tabular}$

Now, plot the points $P(1,6), Q(2,4)$ . The point $B(3,2)$ has already been plotted. Join $P Q$ and $Q B$ to get the graph line $P B$ . Extend it on both ways.