Draw the graph of the pair of equations $2 x + y = 4$ and $2 x - y = 4$ write the vertices of the triangle formed by these lines and the y-axis .also find the area of this triangle.

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Solution

$2 x + y = 4 . . . (i)$
$\Rightarrow y = 4 - 2 x$
If $x = 0, y = 4 - 2 (0) = 4 - 0 = 4$
$x = 1, y = 4 - 2 (1) = 4 - 2 = 2$
$x = 2, y = 4 - 2 (2) = 4 - 4 = 0$
$x$ $0$ $1$ $2$ $3$
$y$ $4$ $2$ $0$ $- 2$
$(i)$ $A$ $B$ $C$ $- 2$
$2 x - y = 4 . . . (i i)$
$\Rightarrow y = 2 x - 4$
If $x = 0, y = 2 (0) - 4 = 0 - 4 = - 4$
$x = 1, y = 2 (1) - 4 = 2 - 4 = - 2$
$x = 2, y = 2 (2) - 4 = 4 - 4 = 0$
$x$ $0$ $1$ $2$ $3$ $4$
$y$ $- 4$ $- 2$ $0$ $2$ $4$
$(i i)$ $E$ $F$ $G$ $H$ $I$
triangle formed by the lines with y -axis $Δ A E C$ coordinates of vertices are $A (0, 4), E (0, - 4)$ and $C (2, 0)$
Area of $Δ A E C = \frac{1}{2} B a s e \times A l t i t u d e$
$= \frac{1}{2} A E \times C O$
$= \frac{1}{2} \times [4 - (- 4)] \times (2 - 0)$
$= \frac{1}{2} \times 8 \times 2 = 8$
$∴$ Area of $Δ A E C =$ Square units.

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