Question

Show that the lines 2x + 5y = 1, x - 3y = 6 and x + 5y + 2 = 0 are concurrent.

Answer 1

We know that two or more lines are said to be concurrent if they intersect at a single point.
We first find the point of intersection of the first two lines.
2x + 5y = 1 ….(1)
x – 3y = 6 ….(2)
Multiplying (2) by 2, we get,
2x – 6y = 12 ….(3)
Subtracting (3) from (1), we get,
11y = -11
y = -1
From (2), x = 6 + 3y = 6 – 3 = 3
So, the point of intersection of the first two lines is (3, -1).
If this point lie on the third line, i.e., x + 5y + 2 = 0, then the given lines will be concurrent.
Substituting x = 3 and y = -1, we have:
L.H.S. = x + 5y + 2 = 3 + 5(-1) + 2 = 5 – 5 = 0 = R.H.S.
Thus, (3, -1) also lie on the third line.
Hence, the given lines are concurrent.