Sir please give me a detailed solution of this question as it was already asked in previous year paper of 2010 in which we have given that 2x-5y+4=0, 2x+y-8=0 and we have to solve the equations graphically and also asked to find the vertices and also the area of the triangle formed by these lines and the y-axis
RS Aggarwal, Class 10, chapter 3A, question no 19
Given equations are
2x−5y+4=0...(1)
2x+y−8=0...(2)
Write y in terms of x for equation (1).
2x−5y+4=0
⇒y=(2x+4)/5
Substitute different values of x in the above equation to get corresponding values of y
For x=−2,y=0
For x=3,y=2
For x=−−7,y=−2
Now plot the points A(−2,0), B(3,2) and C(−7,−2) in the graph paper and join A, B and C to get the graph of 2x−5y+4=0
Similarly, Write y in terms of x for equation (2).
2x+y−8=0
⇒y=8−2x
Substitute different values of x in the above equation to get corresponding values of y
For x=4,y=0
For x=3,y=2
For x=2,y=4
Now plot the points D(4,0), E(3,2) and F(2,4) in the graph paper and join D, E and F to get the graph of 2x+y−8=0
From the graph:
Both the lines intersect each other at point E(3,2) and y-axis at G(0,54) and H(0,8)
[Co-ordinates of point G and H are obtained from graph or put x=0 in equation (1) and (2)]
∴ Area of △HGE=21(base×altitude)
=1/2×(7 1/4)×3 sq.units
=1/2×(29/4)×3 sq.units
=87/8 sq.units
=10.875 sq.units.