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At what angle do line y=Kx + B and line y=Bx+K intersect? 1.

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At what angle do line y=Kx + B and line y=Bx+K intersect? 1. [#permalink] New post 19 Nov 2007, 07:32
At what angle do line y=Kx + B and line y=Bx+K intersect?

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

Please walk me thru this problem.
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 [#permalink] New post 19 Nov 2007, 07:49
Expert's post
C.

angle(kx+b,bx+k)=angle(kx,bx) y=kx+b (k: slope, b: shift)
k,b - characterize slopes of lines.

1. b+k=1
for example, (b,k)=(0,1) ==> angle=45
(b,k)=(1000001,-1000000) ==> angle~0
insuff.

2. b*k=0
for example, (b,k)=(0,1) ==> angle=45
(b,k)=(0,-1000000) ==> angle~90
insuff.

3. b+k=1, b*k=0 ==> (b,k) e {(0,1);(1,0)}
for both cases angle=45
suff.
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 [#permalink] New post 19 Nov 2007, 07:53
Answer is C. Both statements together are sufficient.

Here is a detailed explanation.

Consider statement 1.

B + K =1.

Substituting this equation in the 2 given equations doesn't tell us much.

Statement 1 is insufficient.

Consider statement 2.

B * K =0.
i.e. either b=0 or k=0 or b=0,k=0.

This does not give us specific information about the position of the lines either.

Consider the two statements together.

B + K =1

&

B * K =0.

That means either B = 0 and K = 1

OR

B = 1 and K = 0.

Considering either of the possibilities in the given line equations leaves us with the same two equations.

i.e y = 1 and y = x

y = 1 is a line parallel to the axis at a height of 1.

y=x will have points such as (0,0), (1,1), (2,2),....

Therefore, these two lines can only intersect at one point (1,1).

We now have an isosceles right angled triangle.

i.e. the two lines intersect at an angle of 45 degrees.

Hope this helps.
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 [#permalink] New post 20 Nov 2007, 11:54
good job guys.

OA is C.

slope of 1 is 45 deg
slope of 0 is flat, parallel to x axis
intersect at 45 deg.
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 [#permalink] New post 20 Nov 2007, 12:36
Another way (more useful for arbitrary angles):
easy formula for any general straight lines:

The angle between two lines y = m1x + c1 and y = m2x + c2 is tan inverse of the modulus of : [(m1 - m2)/(1 + m1*m2)]

tan X = ( B - K)/(1 + B*K)

case 1: tan X = ( B - K)/(1 + B*K) ( not sufficient)

case 2: tan X = (B - K) (not sufficient)

combine case 1 and 2:

tan X = (B - K) where K = 0 or B = 0. Therefore, B = 1 or K = 1 respectively.

Hence tan X = 1;(taking the positive angle) X = 45 degrees.
  [#permalink] 20 Nov 2007, 12:36
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