Question
Solve graphically, the following equations.$x + 2y = 4; 3x - 2y = 4$.Take $2 \ cm = 1$ unit on each axis.Also, find the area of the triangle formed by the lines and the x-axis.
✓
Answer
$x + 2y = 4$
$⇒ x = 4 - 2y$
The table of $x + 2y = 4$ is
$3x - 2y = 4$
$\Rightarrow x =\frac{4+2 y}{3}$
The table of $3x - 2y = 4$ is
Now plotting the points on a graph and we get the following required graph:
Therefore the solution of the given system of equations is $(2,1).$
Thus the vertices of the triangle are:
$A(2,1), B\left(\frac{4}{3}, 0\right)$ and $C(4,0)$
$AB ⊥ BC$ and $D ≡ (2,0)$
$AD = 1$ and $BC =2 \frac{2}{3}$ units $=\frac{8}{3}$ units
Area of the triangle ABC $=\frac{1}{2} \times AD \times BC$
$=\frac{1}{2} \times 1 \times \frac{8}{3}$
$=\frac{4}{3} \text { sq}$.units
$=1 \frac{1}{3} \text { sq}$.units
$⇒ x = 4 - 2y$
The table of $x + 2y = 4$ is
| $X$ | $2$ | $- 4$ | $12$ |
| $Y$ | $1$ | $4$ | $- 4$ |
$\Rightarrow x =\frac{4+2 y}{3}$
The table of $3x - 2y = 4$ is
| $X$ | $2$ | $4$ | $6$ |
| $Y$ | $1$ | $4$ | $7$ |
Therefore the solution of the given system of equations is $(2,1).$
Thus the vertices of the triangle are:
$A(2,1), B\left(\frac{4}{3}, 0\right)$ and $C(4,0)$
$AB ⊥ BC$ and $D ≡ (2,0)$
$AD = 1$ and $BC =2 \frac{2}{3}$ units $=\frac{8}{3}$ units
Area of the triangle ABC $=\frac{1}{2} \times AD \times BC$
$=\frac{1}{2} \times 1 \times \frac{8}{3}$
$=\frac{4}{3} \text { sq}$.units
$=1 \frac{1}{3} \text { sq}$.units
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