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by GMATGuruNY » Fri Jan 27, 2012 10:15 am
If lines k and l are perpendicular, they will form four 90 degree angles.
If lines k and l are not perpendicular, they will form two angles less than 90 degrees and two angles greater than 90 degrees.

Question rephrased: Are lines k and l perpendicular?

Many slope questions are best solved by DRAWING.

Statement 1:
Since each line has a positive y-intercept and passes through (3,-2), the two lines cannot be perpendicular:
Image
SUFFICIENT.

Statement 2:
Case 1: Lines k and l are perpendicular, and the two y-intercepts are EQUIDISTANT from (3,-2).
Image
Here, the distance between the y-intercepts = 6.

Case 2: Lines k and l are perpendicular, and the two y-intercepts are NOT EQUIDISTANT from (3,-2).
Image
Here, the distance between the y-intercepts > 6.

Thus, if lines k and l are perpendicular, the MINIMUM distance between the y-intercepts is 6.
Since statement 2 indicates that the distance between the y-intecepts is 5, lines k and l are not perpendicular.
SUFFICIENT.

The correct answer is D.
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by GMATGuruNY » Fri Jan 27, 2012 11:18 am
I received a request for an algebraic approach to statement 2.

Let d = the distance between the y-intercepts of two perpendicular lines that intersect to the right of the y-axis.
Let's assume that neither line has an undefined slope.
Let the equation of the line with the greater y-intercept be y = mx+b.
Then the equation of the line with the smaller y-intercept is y = (-1/m)x + (b-d).

At the point of intersection, the two lines have the same y value:
mx + b = (-1/m)x + (b-d)
mx = (-1/m)x - d
x(m²) = -x - dm
x(m²) + dm + x = 0.

The discriminant of ax² + bx + c = 0 is b²-4ac.
Thus, the discriminant of x(m²) + dm + x = 0 is:
d²-4(x)(x) = d² - 4x².

For a quadratic to have a real solution, its discriminant must be greater than or equal to 0:
d² - 4x² ≥ 0
d² ≥ 4x²
d²/x² ≥ 4
d/x ≥ 2.

In the inequality above:
d is the distance between the y-intercepts.
x is the distance between the y-axis and the point of intersection.
The value of d is at least two times the value of x.

Thus, the distance between the y-intercepts is AT LEAST TWICE the distance between the y-axis and the point of intersection.
This will be true of any two perpendicular lines with defined slopes.

In statement 2, the distance between the y-intercepts (5) is less than twice the distance between the y-axis and (3,-2).
Thus, lines k and l are not perpendicular.
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by Prashant Ranjan » Sun Jan 29, 2012 4:03 am
Pardon Mitch for asking me this.
But if we assume the distance between the two y intercepts of the perpendicular lines to be d and y intercept of one line to be 'b', then shouldn't the y intercept of the other line be 'd-b'?

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by GMATGuruNY » Sun Jan 29, 2012 5:03 am
Prashant Ranjan wrote:Pardon Mitch for asking me this.
But if we assume the distance between the two y intercepts of the perpendicular lines to be d and y intercept of one line to be 'b', then shouldn't the y intercept of the other line be 'd-b'?

Thanks
My explanation indicates that the equation of the line with the GREATER y-intercept is y = mx + b.
Since the other line has a LOWER y-intercept -- and the positive distance between the y-intercepts is d -- the y-intercept of the other line is b-d.
To illustrate:
If the higher y-intercept is b=10, and the distance between the y-intercepts is d=2, the lower y-intercept = b-d = 10-2 = 8.
(Using your logic, the lower y-intercept would be d-b = 2-10 = -8. Here, the distance between the y-intercepts = 10-(-8) = 18, which is not the value of d.)
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