Question

Show that the straight lines x + y – 4 = 0, 3x + 2 = 0 and 3x – 3y + 16 = 0 are concurrent.

Answer 1

The lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, a₃x + b₃y + c₃ = 0 are concurrent if

`|(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)|` = 0

The given lines x + y – 4 = 0, 3x + 0y + 2 = 0, 3x – 3y + 16 = 0

`|(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)| = |(1,1,-4),(3,0,2),(3,-3,16)|`

= 1(0 + 6) – 1(48 – 6) – 4(-9 – 0)

= 6 – (42) + 36

= 42 – 42

= 0

The given lines are concurrent.