Question

Find the equation of line parallel to the y-axis and drawn through the point of intersection of x – 4y + 1 = 0 and 2x + y – 7 = 0.

Options

• x – 1 = 0

• x – 2 = 0

• x – 3 = 0

• x – 4 = 0

Answer 1

x – 3 = 0

Explanation:

Let `l_1 : x - 4y + 1` = 0, `l_2 : 2x + y - 7` = 0

Equation of line passing through the point of intersection of lines `l_1` and `l_2`.

`l_3 : (x - 4y + 1) + k(2x + y - 7)` = 0

`(x + 2kx) + (-4y + ky) + (1 - 7k)` = 0

∴ `(1 + 2k)x + (k - 4)y + (1 - 7k)` = 0  .....(i)

`x/((7k - 1)/(2k + 1)) + y/((7k - 1)/(k - 4))` = 1

As the line `l_3` is parallel to y-axis. Then

`(7k - 1)/(k - 4) = 1/0` ⇒ `k - 4` = 0

∴ `k` = 4

Thus, the equation of line is `(1 + 8) + 0.y + (1 - 28)` = 0

`9x - 27` = 0

`x - 3` = 0