Answer: 2x−y−4=0 …(i) y=2x−4 When x=1,y=2×1−4=2−4=−2 When x=2,y=2×2−4=4−4=0 When x=3,y=2×3−4=6−4=2 x= 1 ,2 ,3 y= −2 ,0 ,2 Mark the above points on graph. Join them. x+y+1=0 …(ii) y=−x−1 When x=0,y=0−1=−1 When x=−2,y=2−1=1 When x=−1,y=1−1=0 x= −2 ,−1 ,0 y= 1 ,0 ,−1 Mark the above points on graph. Join them. It is clear from the graph that the two lines intersect at (1,−2). So the solution of the given equations are x=1 and y=−2. The area of the triangle formed by these lines and Y axis =1/2 ×base×height =1/2×3×1 =1.5sq. units Hence area of the triangle is 1.5sq. units Reply

Answer:2x−y−4=0 …(i)

y=2x−4

When x=1,y=2×1−4=2−4=−2

When x=2,y=2×2−4=4−4=0

When x=3,y=2×3−4=6−4=2

x= 1 ,2 ,3

y= −2 ,0 ,2

Mark the above points on graph. Join them.

x+y+1=0 …(ii)

y=−x−1

When x=0,y=0−1=−1

When x=−2,y=2−1=1

When x=−1,y=1−1=0

x= −2 ,−1 ,0

y= 1 ,0 ,−1

Mark the above points on graph. Join them.

It is clear from the graph that the two lines intersect at (1,−2).

So the solution of the given equations are x=1 and y=−2.

The area of the triangle formed by these lines and Y axis =1/2 ×base×height

=1/2×3×1

=1.5sq. units

Hence area of the triangle is 1.5sq. units