Question

Draw the graphs of the following equations on the same graph paper:
$$\begin{gathered} 2x+y=2 \\ 2x+y=6 \end{gathered}$$
Find the coordinates of the vertices of the trapezium formed by these lines. Also, find the area of the trapezium so formed.

Answer 1

From the first equation, write y in terms of x
$y=2-2x.....\left(i\right)$
Substitute different values of x in (i) to get different values of y
$$\begin{gathered} \mathrm{For}x=0,y=2-0=2 \\ \mathrm{For}x=1,y=2-2=0 \\ \mathrm{For}x=2,y=2-4=-2 \end{gathered}$$
Thus, the table for the first equation (2x + y = 2) is

x	0	1	2
y	2	0	−2

Now, plot the points A(0,2), B(1,0) and C(2,−2) on a graph paper and join
A, B and C to get the graph of 2x + y = 2.
From the second equation, write y in terms of x
$y=6-2x.....\left(ii\right)$
Now, substitute different values of x in (ii) to get different values of y
$$\begin{gathered} \mathrm{For}x=0,y=6-0=6 \\ \mathrm{For}x=1,y=6-2=4 \\ \mathrm{For}x=3,y=6-6=0 \end{gathered}$$
So, the table for the second equation (2x + y = 6 ) is

x	0	1	3
y	6	4	0

Now, plot the points D(0,6), E(1,4) and F(3,0)on the same graph paper and join
D, E and F to get the graph of 2x + y = 6.



From the graph it is clear that, the given lines do not intersect at all when produced. So, these lines are
parallel to each other and therefore, the quadrilateral DABF is a trapezium. The vertices of the
required trapezium are D(0,6), A(0,2), B(1,0) and F(3,0).
Now,
$$\begin{gathered} \mathrm{Area}\left(TrapeziumDABF\right)=\mathrm{Area}\left(∆DOF\right)-\mathrm{Area}\left(∆AOB\right) \\ =\frac{1}{2}\times 3\times 6-\frac{1}{2}\times 1\times 2 \\ =9-1 \\ =8\mathrm{sq}.\mathrm{units} \end{gathered}$$
Hence, the area of the rquired trapezium is 8 sq. units.