abhi332 wrote:On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What is the distance between point (0, 0) and the closest vertex of the square?
(A) 1/sqrt (2)
(B) 1
(C) sqrt (2)
(D) sqrt (3)
(E) 2*sqrt (3)
[spoiler]
OA : C[/spoiler]
Equation of the diagonal x-6/(6) = y-2/(-4)
-4x+24 = 6y -12
2x+3y -18 =0
The other diagonal will be perpendicular to the first and bisects the first at (3,4)
the equation is 3x-2y +c =0
since it passes through (3,4) 3*3-2*4 +c =0
c = 1
2x-3y= -1
Half the length of the diagonal = sqrt(3^2+2^2) (distance from (3,4) to (0,6))
Now we have to find a point which is at sqrt(3^2+2^2) distance from (3,4) on the line 2x-3y= -1
(1,1) and (4,7) fits the bill of which, (1,1) is closer to the origin
Distance of (1,1) to the origin = sqrt(1^2+1^2) =[spoiler]sqrt(2)[/spoiler]
[spoiler]This is neither an elegant method nor an easy method![/spoiler]
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