Show that the lines 2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are c

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

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

Answer»

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.