my approach is just visualize each line in mind
( i know its easier said than done...but considering 2 mins time....i think this is the fastest way)
refer image attached.......Now the task is to find min distance from a line in options to the point (1,0).......this can also be done by a formula from a geometry chapter
First visualise point (1,0)...now to calculate min distance....here its very helpful to know a property.....in 45-45-90 triangle.....sides are in 1:1:srt(2) proportion....now apply this in following options
1) y=x....since y=x.....aangle of line wrto x axis = 45 ( shown in fig)....we need perpendicular distance hence another angle = 90 deg....thus it becomes a triangle of 45-45-90....hence hypoteneous is 1...now equate the proportion as 1:1:srt(2) = x:x:1.....this gives x = 1/srt(2)min distance = 1/srt(2)
2)y=1....min distance = 1
3)y+x=3......min distance = srt(2).......apply the same logic as of option 1
4)x=2.....min distance = 1
5)x+y=-1....min distance = srt(2).......apply the same logic as of option 1
OA A _________________
Bhushan S.
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