Question

Prove that the lines 4x + 3y = 10, 3x - 4y = - 5 and 5x + y = 7 are concurrent.

Answer 1

Given lines are

4x + 3y = 10

⇒ 4x + 3y - 10 = 0 ⇒ 3x - 4y = - 5

⇒ 3x- 4y + 5 = 0 ⇒ 5x + y = 7

⇒ 5x + y - 7 = 0  

The condition for the given lines to be concurrent is

`|(4,3,-10),(3,-4,5),(5,1,-7)|` = 0

Consider `|(4,3,-10),(3,-4,5),(5,1,-7)|`

`=> 4|(-4,5),(1,-7)| - 3|(3,5),(5,-7)| - 10|(3,-4),(5,1)|`    ...[Expanding along R₁]

⇒ 4(28 - 5) - 3(- 21 - 25) - 10(3 + 20)

⇒ 4(23) - 3(- 46) - 10(23)

⇒ 92 + 138 - 230

⇒ 230 - 230 = 0

Hence, the given lines are concurrent.