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