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प्रश्न
Solve graphically the system of linear equations:
4x − 3y + 4 = 0
4x + 3y − 20 = 0
Find the area bounded by these lines and x-axis.
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उत्तर
The given equations are
4x − 3y + 4 = 0 ...........(i)
4x + 3y − 20 = 0 .............(ii)
Putting x = 0 in equation (i) we get
=> 4 xx 0 - 3y = -4
=> y = 4/3
x = 0, y = 4/3
Putting y = 0 in equation (i) we get
=> 4x - 3 xx 0 = -4
=> x = -1, y = 0
Use the following table to draw the graph.
| x | 0 | -1 |
| y | 4/3 | 0 |
The graph of (i) can be obtained by plotting the points (0, 4/3), (−1, 0).
4x + 3y = 20 ..........(ii)
Putting x= 0 in equation (ii) we get
`=> 4 xx 0 + 3y = 20`
=> y = 20/3
x = 0, y = 20/3
Putting y= 0 in equation (ii) we get
`=> 4x + 3 xx 0 = 20`
=> x = 5
x = 5, y = 0
Use the following table to draw the graph.
| x | 0 | 5 |
| y | 20/3 | 0 |
Draw the graph by plotting the two points from table.

The two lines intersect at P(2,4)
Hence x = 2, y = 2 is the solution of the given equations.
Now,
Required area = Area of PBD
Required area = 1/2 (base x height)
Required area = 1/2 (BD x PM)
Required area = 1/2 (6 x 4)
Hence, the area = 12 sq. units
