Question

Solve graphically the system of equations
x – y – 3 = 0
2x – 3y – 4 = 0.
Find the coordinates of the vertices of the triangle formed by these two lines and the y-axis.

Answer 1

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

x	-4	-1	2
y	4	2	0

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 (ΔEAF) =`1/2xx AF xxEM`

                    = `1/2 xx 5 xx2`

                   = 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.