Draw the graphs of the lines 2x + y = 6 and 2x – y + 2 = 0. Shade the region bounded by these lines and the x-axis.

Solution:

$2 x+y=6$

$\Rightarrow y=-2 x+6$

When $x=0, y=-2 \times 0+6=0+6=6$

When $x=1, y=-2 \times 1+6=-2+6=4$

When $x=2, y=-2 \times 2+6=-4+6=2$

Thus, the points on the line 2x + y = 6 are as given in the following table:

Plotting the points (0, 6), (1, 4) and (2, 2) and drawing a line passing through these points, we obtain the graph of of the line 2x + y = 6.

$2 x-y+2=0$

$\Rightarrow y=2 x+2$

When $x=0, y=2 \times 0+2=0+2=2$

When $x=1, y=2 \times 1+2=2+2=4$

 

When $x=-1, y=2 \times(-1)+2=-2+2=0$

Thus, the points on the line 2x – y + 2 = 0 are as given in the following table:

Plotting the points (0, 2), (1, 4) and (–1, 0) and drawing a line passing through these points, we obtain the graph of of the line 2x – y + 2 = 0.

The shaded region represents the area bounded by the lines 2x + y = 6, 2x – y + 2 = 0 and the x-axis. This represents a triangle.

It can be seen that the lines intersect at the point C(1, 4). Draw CD perpendicular from C on the x-axis.

Height = CD = 4 units

Base = AB = 4 units

$\therefore$ Area of the shaded region = Area of $\Delta \mathrm{ABC}=\frac{1}{2} \times \mathrm{AB} \times \mathrm{CD}=\frac{1}{2} \times 4 \times 4=8$ square units