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line y=Kx+B and line y=Bx+K intersect?

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line y=Kx+B and line y=Bx+K intersect?

by PGMAT » Sat Aug 11, 2012 5:37 am
At what angle do line y=Kx+B and line y=Bx+K intersect?

1. B+K=1
2. B*K=0

[spoiler]OA is C. the lines intersect at 45 degrees.
I arrived at A. Can some one explain what I am missing.
I solved both equations, and came up with x=1, y=K+B. From 1) B+K=1, which means (1,1) and the angle is 45 degrees. So how is A alone not sufficient? [/spoiler]

Thanks.
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by niketdoshi123 » Sat Aug 11, 2012 6:54 am
PGMAT wrote:At what angle do line y=Kx+B and line y=Bx+K intersect?

1. B+K=1
2. B*K=0

[spoiler]OA is C. the lines intersect at 45 degrees.
I arrived at A. Can some one explain what I am missing.
I solved both equations, and came up with x=1, y=K+B. From 1) B+K=1, which means (1,1) and the angle is 45 degrees. So how is A alone not sufficient? [/spoiler]

Thanks.
As shown in the diagram below, the two lines intersect at (1,1) but clearly the angle at which they intersect is not 45°.

Hence statement 1 is sufficient to find the point of intersection of the two lines but is insufficient to find the angle of their intersection.

combining both the statements

B + K = 1
=> B = 1 - K

B*K = 0
=> K* ( 1 - K) = 0
Hence K = 0 or K = 1
for K = 0 , B = 1
and for K = 1 , B = 0

Let K = 1 and B = 0
line 1: y = Kx + B
Here slope , K = 1 . hence the line is 45° to the y-axis
line 2: y = Bx + K
Here the slope , B = 0 . hence the line is 90° to the y-axis

Therefore the lines intersect at 45°angle. Hence sufficient

K = 0 and B = 1 will also give the same result.

The correct answer is C
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