From the first equation, write y in terms of x
y = \(\frac{2x+4}5\).....(i)
Substitute different values of x in (i) to get different values of y
For x = -2, y = \(\frac{-4+4}5 = 0\)
For x = 0, y = \(\frac{0+4}5 = \frac{4}5\)
For x = 3, y = \(\frac{6+4}5 = 2\)
Thus, the table for the first equation (2x – 5y + 4 = 0) is
| x |
-2 |
0 |
3 |
| y |
0 |
\(\frac{4}5\) |
2 |
Now, plot the points A(-2, 0), B \((0,\frac{4}5)\) and C(3, 2) on a graph paper and join A, B and C to get the graph of 2x – 5y + 4 = 0.
From the second equation, write y in terms of x
y = 8 – 2x …….(ii)
Now, substitute different values of x in (ii) to get different values of y
For x = 0, y = 8 – 0 = 8
For x = 2, y = 8 – 4 = 3
For x = 4, y = 8 – 8 = 0
So, the table for the second equation (2x – 5y + 4 = 0) is
Now, plot the points D(0, 8), E(2, 4) and F(4, 0) on the same graph paper and join D, E and F to get the graph of 2x + y - 8 = 0.
From the graph, it is clear that, the given lines intersect at (3, 2).
So, the solution of the given system of equation is (3, 2).
The vertices of the triangle formed by the system of equations and y-axis are (0, 8), (0, 4 5 ) and (3, 2).
Draw a perpendicular from point C on the y-axis. So,
Area (∆DBC) = \(\frac{1}2\) × DB × CM
= \(\frac{1}2\) x \((8-\frac{4}5)\) x 3
= \(\frac{54}5\) sq. units
Hence, the vertices of the triangle are (0, 8), \((0,\frac{4}5)\) and (3, 2) and its area is\(\frac{54}5\) sq. units
