From the first equation, write y in terms of x
y = x + 3 …….(i)
Substitute different values of x in (i) to get different values of y
For x = -3, y = -3 + 3 = 0
For x = -1, y = -1 + 3 = 2
For x = 1, y = 1 + 3 = 4
Thus, the table for the first equation (x - y + 3 = 0) is
Now, plot the points A(-3, 0), B(-1, 2) and C(1, 4) on a graph paper and join A, B and C to get the graph of x - y + 3 = 0.
From the second equation, write y in terms of x
y = \(\frac{4-2x}3\)......(ii)
Now, substitute different values of x in (ii) to get different values of y
For x = -4, y = \(\frac{4+8}3\) = 4
For x = -1, y = \(\frac{4+12}3\) = 2
For x = 2, y = \(\frac{4-4}3\) = 0
So, the table for the second equation (2x + 3y - 4 = 0) is
Now, plot the points D(-4, 4), E(-1, 2) and F(2, 0) on the same graph paper and join D, E and F to get the graph of 2x + 3y - 4 = 0
From the graph, it is clear that, the given lines intersect at (-1, 2).
So, the solution of the given system of equation is (-1, 2).
The vertices of the triangle formed by the given lines and the x-axis are (-3, 0), (-1, 2) and (2, 0).
Now, draw a perpendicular from the intersection point E on the x-axis. So,
Area (∆EAF) = \(\frac{1}2\) × AF × EM
= \(\frac{1}2\) × 5 × 2
= 5 sq. units
Hence, the vertices of the triangle formed by the given lines and the x-axis are (-3, 0), (-1, 2) and (2, 0) and its area is 5 sq. units.
