The larger of the two angles made with x-axis of a straight line drawn through (1,2) so that it intersects x+y= 4 at a distance (√6 ) / 3 from (1,2) is : -(a) 15 º (b) 60 º (c) 75 º (d) 105 º

let slope of line is m then eq of line will be

L : (y-2) = (x-1)m

now this line intersects x+y = 4 , let point os intersection be A then

A = { (m+2/m+1) , (3m+2/m+1) }

now distance bw (1,2) & this point A is given root6 / 3 so

by using distance formula & after equating to root6/3

we get a quadratic in m

m² - 4m + 1 = 0

m = tan@ =  2 + roo3 or 2-root3

@ = 75    from +ve x axis  so

it can form maximum 105^o from -ve x axis .....

option d is correct ,

approve if u like my ans

let slope of line is m then eq of line will be

L : (y-2) = (x-1)m

 now this line intersects x+y = 4 , let point os intersection be A then

A = { (m+2/m+1) , (3m+2/m+1) }

 now distance bw (1,2) & this point A is given root6 / 3 so by using distance formula & after equating to root6/3 we get a quadratic in m

m² - 4m + 1 = 0

m = tan@ =  2 + roo3 or 2-root3

@ = 75    from +ve x axis  s, it can form maximum 105^o from -ve x axis .....

D is d r8 ansr.