Draw the graphs representing the equations 4x + 3y = 24 and 3y = 4x + 24 on the same graph paper. Write the co – ordinates of the point of intersection of these lines and find the area of the triangle formed by these lines and the x – axis.

Sol. For 4x + 3y = 24,

x	0	3	6
y	8	4	0
(x , y)	(0, 8)	(3, 4)	(6, 0)

Put, x = 0, we get y = 8

Put x = 3 we get y = 4 and

put x = 6 we get y = 0

For 3y = 4x + 24 ,

x	- 6	-3	0
y	0	4	8
(x , y)	(-6,  0)	(-3, 4)	(0,8)

Put x = - 6 we get y = 0

Put x = -3 we get y = 4

Put x = 0 we get y = 8

Now, we have ∆ ABC,

With Base BC = 12 units, Height AO = 8 units,

We know that, A (∆ ABC) = ½ (Base) (Height)

∴ A(∆ ABC) = ½ (BC)(AO)

∴ A(∆ ABC) = ½ (12)(8)

∴ A(∆ ABC) =  6 × 8

∴ A(∆ ABC) = 48 sq. units.