# Draw the graphs of the following equations on the same graph paper:2x + y = 2, 2x + y = 6.Find the coordinates of the vertices of the trapezium formed by these lines. Also, find the area of the trapezium so formed.

For equation, 2x + y = 2

First, take x = 0 and find the value of y.

Then, take y = 0 and find the value of x.

Now similarly solve for equation, 2x + y = 6

Plot the values in a graph and find the intersecting point for the solution.

Since, the line 2x + y = 6 cuts the line y - axis at A(0,6) and x - axis at B(3,0)

& the line 2x + y = 2 cuts the x - axis at C(1,0) and y - axis at D(0,2).

Thus, it is clear from the graph that ABCD forms a trapezium.

And the coordinates joining this trapezium are (0,6),(3,0),(1,0) and (0,2).

We can find the area of trapezium ABCD.

The formula to calculate area of a trapezium ABCD is:

Area(trap. ABCD) = Area(∆OAB) – Area(∆OCD)

= (1/2 × 3 × 6) – (1/2 × 1 × 2)

[ base(∆OAB) = 3 units & height(∆OAB) = 6 units

= 9 - 1 base(∆OCD) = 1 units & height(∆OCD) = 2units]

= 8 sq. units

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