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`

Answer

N/a

Explanation

Let of the line be m. As, the line passes through the point A(1,2).
`therefore` Equation of line is `y-2-m=0`
`mx-y+2-m=0`
and `x+y-4=0`
`(x)/(4-2+m)=(y)/(2-m+4m)=(1)/(1+m)`
`rArr(x)/(2+m)=(y)/(3m+2)=(1)/(1+m)`
`rArrx=(2+m)/(1+m)`
`y=(3m+2)/(1+m)`
So, the point of intersection is B `((m+2)/(m+1),(3m+2)/(m+1))`
Now, `AB^2((m+2)/(m+1)-1)^2+((3m+2)/(m+1)-2)^2`
`because AB=sqrt(6)/(3)`
`therefore((m+2-m-1)/(m+1))^2+((3m+2-2m-2)/(m+1))^2=(6)/(9)`
`rArr((1)/(m+1))^2+((m)/(m+1))^2(6)/(9)`
`rArr(1+m^2)/(1+m)^2=(6)/(9)`
`rArr(1+m^2)/(1+m^2+2m)=(6)/(9)`
`rArr9+9m^2=6+6m^2+12m`
`rArr3m^2=12m+3=0`
`rArrm^2-4m+1=0`
`therefore m=(4pmsqrt(16-4))/(2)`
`=2pmsqrt(3)`
`=2+sqrt3` or `2-sqrt3`
`thereforetheta=75^@` or ` 15^@`