Question

Prove that the lines 3x + y – 14 = 0, x – 2y = 0 and 3x – 8y + 4 = 0 are concurrent

Answer 1

Given equation of lines are

 3x + y – 14 = 0     …(1)

 x – 2y = 0             …(2) 

3x – 8y + 4 = 0      …(3) 

Solving (1) and (2), (1) – (3) × (2) ⇒ y + 6y – 14 = 0 ⇒ y = 2 

Putting in x – 2y = 0, x = 4 

∴ (4, 2) is the common point of (4, 2) substituting this point in (3) 

L.H.S = 3.4 – 8.2 + 4 = 0 = R.H.S 

∴ The point of intersection of (1) and (2) satisfies the equation (3), so the given three points are the equation (3), so the given three points are concurrent.