Question

Draw the graphs of the linear equations 4x − 3y + 4 = 0 and 4x + 3y − 20 = 0. Find the area
bounded by these lines and x-axis.

Answer 1

We have 

     4x - 3y +  4 = 0 

⇒  4x - 3y = 4 

⇒ ` x = ( 3y  - 4 ) / 4 `

Putting   y = 0 . we get ` ( 3 xx 0 - 4) / 4  = -1`

Putting  y  = 4 , we get ` ( 3 xx 4- 4) / 4  = - 2`

Thus, we have the following table for the p table for the points on the line   4x - 3y  + 4 = 0 

x	- 1	2
y	0	4

 we have 

      4x + 3y - 20 = 0 

⇒   4x = 20 - 3y 

⇒ ` x = ( 20 - 3y )/ 4`

Putting  y = 0 , we get  ` x  = (20 - 3 xx 0) /4 = 5  `

Puttiing  y = 4  , we get ` x = ( 20 - 3 xx 4 ) / 4 = 2 `

Thus, we have the following table for the p table for the points on the line 4x  - 3y - 20 = 0 

x	0	2
y	0	4

\

 

Clearly, two lines intersect at A ( 2 , 4 ) 
The graph of the lines  4x - 3y + 4 = 0   and  4x  + 3y - 20 = 0  intersect with y - axis at
 a + B (- 1 , 0 ) and  c ( 5 , 0 )respectively

∴ Area of `Δ ABC  = 1/ 2  [  Base xx height ]`

  =` 1 / 2  ( BC xx AB )`

  =` 1 / 2  ( 6 xx 4)`

   = ` 3 xx 4 `

12 sq .units 

∴ Area of Δ ABC  = 12 sq .units