Answer :
Let us call 3x+y-12=0 Line1 and x-3y+6=0 Line2
Points on line 1 are (0,12), (3,3),(4,0)and on line 2 are (0,2),(3,3),(-6,0).
Draw these two lines through respective points .These will be found to intersect at (3,3). Let us call point (3,3) as H.
Line 1 will intersect X-axis at (4,0)[ Let us call point (4,0) as B. And Line 2 will intersect X-axis at (-6,0) [Let us call this point as A.
If you draw line parallel to y-axis from point (3,3), it will be perpendicular to X-axis and the point where it meets x-axis be called Q.
Then the triangle HAB will be the shaded triangular area whose base = 1o and height = HT = 3. Hence area of shaded area = (1/2)xBase x height = (1/2)x10x3 = 15 sq units
Points on line 1 are (0,12), (3,3),(4,0)and on line 2 are (0,2),(3,3),(-6,0).
Draw these two lines through respective points .These will be found to intersect at (3,3). Let us call point (3,3) as H.
Line 1 will intersect X-axis at (4,0)[ Let us call point (4,0) as B. And Line 2 will intersect X-axis at (-6,0) [Let us call this point as A.
If you draw line parallel to y-axis from point (3,3), it will be perpendicular to X-axis and the point where it meets x-axis be called Q.
Then the triangle HAB will be the shaded triangular area whose base = 1o and height = HT = 3. Hence area of shaded area = (1/2)xBase x height = (1/2)x10x3 = 15 sq units