Question

solve the equation graphically :2x+y=2;2y-x=4 find the area of the triangle formed by the two lines and x-axis
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Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of linear equations are

$\rm :\longmapsto\:2x + y = 2 - - - - (1)$

and

$\rm :\longmapsto\:2y - x= 4 - - - - (2)$

Consider,

$\rm :\longmapsto\:2x + y = 2$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:2 \times 0+ y = 2$

$\rm :\longmapsto\: 0+ y = 2$

$\bf\implies \:y = 2$

Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\:2x + 0 = 2$

$\rm :\longmapsto\:2x = 2$

$\bf\implies \:x = 1$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 2 \\ \\ \sf 1 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 2) & (1 , 0)

➢ See the attachment graph. Red line graph

Consider,

$\rm :\longmapsto\:2y - x= 4$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:2y - 0= 4$

$\rm :\longmapsto\:2y = 4$

$\bf\implies \:y = 2$

Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\:2 \times 0 - x= 4$

$\rm :\longmapsto\:0 - x= 4$

$\rm :\longmapsto\: - x= 4$

$\bf\implies \:x = - 4$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 2 \\ \\ \sf - 4 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 2) & (- 4 , 0)

➢ See the attachment graph. Blue line graph

Now,

From graph, the area of triangle ABC formed by the lines with x - axis is

$\rm :\longmapsto\:Area_{triangle ABC)} = \dfrac{1}{2} \times 5 \times 2$

$\bf :\longmapsto\:Area_{triangle ABC)} = 5 \: square \: units$