three lines x+3y=6 ,2x-3y=12 and x=0 are enclosing a beautiful triangular park.find the points of intersection of the lines graphically and the area of the park if all the measurements are in km.

Answer :

We have Three lines As :  x  +  3y  = 6  , 2x  - 3y  =  12  And x = 0

First we form a table for different values of x and y for all three equations As  :
 

x  +  3y  = 6		2x  - 3y  =  12		x = 0
x	y	x	y
0	2	0	-4	y -axis
-3	3	-3	-6
-6	4	-6	-8
3	1	3	-2
6	0	6	0

Now we plot these points on x  - y plane , As  :


Here We have triangular park ABC  And point of intersection we get graphically AS :

A (  6 , 0 )  , B (  0 , 2 )  And  C ( 0 , - 4 )

And Area of triangle we know  = $\frac{1}{2}$ ( Base ) ( Height )

So , For Triangle ABC  Base  BC  =  6  km  And Height ( y axis to point A ) =  6 km 
SO,

Area of triangular park ABC  =  $\frac{1}{2}$ ( 6 ) ( 6 )  =  ( 3 )( 6 )  =  18  km²  ( Ans  )