Question

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

Answer 1

To prove given lines

2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent, we have to prove that

∴ They will pass through the same point

Now 2x + 5y = 1  ...(i)

x – 3y = 6  ...(ii)

Multiply (i) by 3 and (ii) by 5, we get:

6x + 15y = 3

5x – 15y = 30

On Adding we get:

11x = 33

`\implies x = 33/11 = 3`

From (ii), we have

x – 3y = 6

`\implies` 3 – 3y = 6

`\implies` –3y = 6 – 3 = 3

`\implies y = 3/(-3) = -1`

∴ Point of intersection of first two lines is (3, –1)

Substituting the values in third line

x + 5y + 2 = 0

L.H.S. = x + 5y + 2

= 3 + 5(–1) + 2

= 3 – 5 + 2

= 5 – 5

= 0 = R.H.S.

Hence the given three lines are concurrent.