Solve graphically the system of equations 2x-3y+4=0 x+2y-5=0 Find the coordinates of the vertices of the triangle formed by these two lines and the $\mathrm{x}$-axis.

Home » Solve graphically the system of equations 2x-3y+4=0 x+2y-5=0 Find the coordinates of the vertices of the triangle formed by these two lines and the -axis.

Solve graphically the system of equations
2x-3y+4=0
x+2y-5=0 Find the coordinates of the vertices of the triangle formed by these two lines and the $\mathrm{x}$ -axis.

Solve graphically the system of equations
2x-3y+4=0
x+2y-5=0 Find the coordinates of the vertices of the triangle formed by these two lines and the $\mathrm{x}$ -axis.

Solution:

From the first eq., write y in terms of $x$
$\mathrm{y}=\frac{2 x+4}{3}\dots \dots(i)$
Substituting different values of $x$ in eq.(i) to get different values of $y$
For $x=-2, y=\frac{-4+4}{3}=0$
For $x=1, y=\frac{2+4}{3}=2$
For $x=4, y=\frac{8+4}{3}=4$

Thus, the table for the first equation $(2 x-3 y+4=0$ ) is

$\begin{tabular}{|r|l|r|r|} \hline $\mathrm{x}$ & $-2$ & 1 & 4 \\ \hline $\mathrm{y}$ & 0 & 2 & 4 \\ \hline \end{tabular}$

Now, plot the points $\mathrm{A}(-2,0), \mathrm{B}(1,2)$ and $\mathrm{C}(4,4)$ on a graph paper and join $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ to get the graph of $2 x-3 y+4=0$ .
From the second eq., write y in terms of $x$
$\mathrm{y}=\frac{5-x}{2}\dots \dots(ii)$
Now, substitute different values of $\mathrm{x}$ in (ii) to get different values of y

For $x=-3, y=\frac{5+3}{2}=4$
For $x=1, y=\frac{5-1}{2}=2$
For $x=5, y=\frac{5-5}{2}=0$
So, the table for the second equation $(x+2 y-5=0)$ is

$\begin{tabular}{|r|l|r|r|} \hline $\mathrm{x}$ & $-3$ & 1 & 5 \\ \hline $\mathrm{y}$ & 4 & 2 & 0 \\ \hline \end{tabular}$

Now, plot the points $\mathrm{D}(-3,4), \mathrm{B}(1,2)$ and $\mathrm{F}(5,0)$ on the same graph paper and join $\mathrm{D}, \mathrm{E}$ and $F$ to get the graph of $x+2 y-5=0$

From the graph, it is clear that, the given lines intersect at $(1,2)$ .
Therefore, the solution of the given system of equation is $(1,2)$ .

From the graph, the vertices of the triangle formed by the given lines and the $\mathrm{x}$ -axis are $(-2$ , $0),(1,2)$ and $(5,0)$
Now, draw a perpendicular from the intersection point $\mathrm{B}$ on the $\mathrm{x}$ -axis. So,
Area $(\Delta \mathrm{BAF})=\frac{1}{2} \times \mathrm{AF} \times \mathrm{BM}$
$\begin{array}{l} =\frac{1}{2} \times 7 \times 2 \\ =7 \text { sq. units } \end{array}$
As a result, the vertices of the triangle formed by the given lines and the $\mathrm{x}$ -axis are $(-2,0),(1,2)$ and $(5,0)$ and the area of the triangle is 7 sq. units.