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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 105o 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 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
@ = 75 from +ve x axis s, it can form maximum 105o from -ve x axis .....
D is d r8 ansr.
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