Find the point of intersection of the lines 4x + 3y = 1 and 3x - y + 9 = 0. If this point lies on the line (2k - 1)x - 2y = 4; find the value of k.

Solving,

4x + 3y = 1 ………(1)

3x - y + 9 = 0 ……….(2)

⇒ y = 3x + 9

Substituting value of y in equation 1 we get,

⇒ 4x + 3(3x + 9) = 1

⇒ 4x + 9x + 27 = 1

⇒ 13x = -26

⇒ x = $\dfrac{-26}{13}$

⇒ x = -2.

Substituting x = -2, in y = 3x + 9 we get,

⇒ y = 3(-2) + 9 = -6 + 9 = 3.

Point of intersection = (-2, 3).

Given, (-2, 3) lies on the line (2k - 1)x - 2y = 4

∴ (2k - 1)(-2) - 2 × 3 = 4

⇒ -4k + 2 - 6 = 4

⇒ -4k - 4 = 4

⇒ 4k = -8

⇒ k = $\dfrac{-8}{4}$

⇒ k = -2.

Hence, point of intersection = (-2, 3) and k = -2.