Angle made with the x-axis by a straight line drawn through (1, 2) so that it intersects `x+y=4` at a distance `(sqrt(6))/3` from (1, 2) is `105^0` (b) `75^0` (c) `60^0` (d) `15^0`

Options

A. `105^(@)`
B. `75^(@)`
C. `60^(@)`
D. `15^(@)`

Answer

B, D

Explanation

Let the angle be `theta`. Then, the equation of the given line is
`(x-1)/("cos" theta) = (y-2)/("sin" theta) " " (i)`
The coordinates of a point (i) a distance `sqrt(6)//3` from (1,2) are `(1+sqrt(6)3 "cos " theta, 2+sqrt(6)3 " sin " theta)`. This point lies on x+y=4. Therefore,
`1+(sqrt(6))/(3)"cos " theta + 2+(sqrt(6))/(3) "sin" theta = 4`
`"or cos " theta + "sin" theta = sqrt((3)/(2))`
`"or "(1)/(sqrt(2)) " cos " theta + (1)/(sqrt(2))"sin" theta = sqrt((3)/(2))`
`"or cos"(theta-(pi)/(4)) = "cos" (+-(pi)/(6))`
`"or "theta-(pi)/(4) = +-(pi)/(6)`
`i.e., theta = 75^(@) " or " theta = 15^(@)`