Question

Draw the graph of 2x+y=6 and 2x-y+ 2=0. Shade the region bounded by these lines and x-axis. Find the area of the shaded region.

Answer 1

We have,

2x + y = 6 \(\Rightarrow\)  y = 6 - 2x

When x = 0, we have 

y = 6 - 2 x 0 = 6

When x = 3, we have

y = 6 - 2 x 3 = 0

When x = 2, we have

y = 6 - 2 x 2 = 2

Thus, we get the following table:

x	0	3	2
y	6	0	2

Now, we plot the points A(0,6), B(3,0) and C(2,2) on the graph paper. We join A, B and C and extend it on both sides to obtain the graph of the equation 2x + y = 6.

We have,

2x - y + 2 = 0 \(\Rightarrow\) y = 2x + 2

When x = 0, we have

y = 2 x 0 + 2 = 2

When x = -1, we have

y = 2 x (-1) + 2 = 0

When x = 1, we have

y = 2 x 1 + 2 = 4

Thus, we have the following table:

x	0	-1	1
y	2	0	4

Now, we plot the points D(0,2), E(-1,0) and F(1,4) on the same graph paper. We join D,E and F and extend it on the both sides to obtain the graph of the equation 2x - y + 2 = 0.

It is evident from the graph that the two lines intersect at point F(1,4). The area enclosed by the given lines and x-axis is shown in Fig. above

Thus, x = 1, y = 4 is the solution of the given system of equations. Draw FM perpendicular from F on x-axis.

Clearly, we have

FM = y-coordinate of point F(1,4) = 4 and BE = 4

\(\therefore\) Area of the shaded region = Area of \(\triangle\)FBE

\(\Rightarrow\) Area of the shaded region = \(\frac{1}{2}\)(Base x Height) = \(\frac{1}{2}\)(BE x FM)

= \(\big(\frac{1}{2}\times4\times4\big)\)sq. units = 8 sq. units.

Answer 2

The correct answer is

Answer 3

ABC is the region bounded by the given lines and the x-axis.

The solution of the given pair of equations is the point where the two lines meet which is (1, 4).

Thus, x = 1 and y = 4.

Area of shaded part = Area of \(\triangle\)ABC

= 1/2 x BC x AD

= 1/2 x 4 x 4

= 8 sq. units