GMAT math question - please help!?

Question

In the xy-coordinate plane, line L and K intersect at point (4,3). Is the product of their slopes negative?

(1) The product of the x-intercepts of lines L and K is positive
(2) The product of the y-intercepts of lines L and K is negative

You must use both statements to find the answer. I have no idea how to solve this, please help!

Thank you!

Answer 1

To answer this, a simple graph, just an intersecting horzontal and vertical line, and then put a point in the upper right quadrant (it does not matter that the point is (4,3), just that both coordinates are positive). Now notice that if a line with negative slope passes through your point, then it must have positive x and y intercepts (a line with both negative intercepts will also have negative slope, but cannot pass through (4,3)). If a lines with positive slope passes through your point then either it has postive x intercept and negative y intercept (if it passes below the origin (0,0)), or negative x intercept and positive y intercept (if it passes above the origin). Facts (1) and (2) tell you that you have one line with positive x and y intercepts and one with postive x intercept and negative y intercept, so one of the lines has positive slope and the other has negative slope, so the product of the slopes must be negative!

Hope that helps!

Answer 2

Sketch a quick graph and draw in the point at (4,3).

In the diagram, you can draw 3 different representative lines for either L or K that pass through (4,3) that have both an x-intercept and a y-intercept (horizontal and vertical lines don't qualify, since they lack one intercept or the other).

Line 1: +ve x-intercept, -ve y-intercept; the slope is +ve.
Line 2: -ve x-intercept, +ve y-intercept; the slope is +ve.
Line 3: +ve x-intercept, +ve y-intercept; the slope is -ve.

L and K can't be both of the same type, since the intercept products would both be +ve (we'd have either a +ve,+ve combo or a -ve, -ve combo, with the product being +ve in both cases.) This contradicts statement 2 in the problem.

Since the product of the x-intercepts has to be +ve, we can eliminate Line 2 as a candidate.

A Line 1, Line 3 combo is the only combo that meets the criteria: we get a +ve x-intercept product and a -ve y-intercept product.

Since Line 1 slope is +ve and Line 3 slope is -ve, the product will be -ve.

Answer 3

I think this is easiest drawn.
Let's "look" at the cases.
Suppose both slopes are positive.
Supposed both lines cross the x axis to the right of the y axis (this means the x intercepts are both positive, and their product positive).
Then, both lines MUST cross the y axis below zero, which makes (2) impossible.
Suppose, then that both lines cross the x axis to the LEFT of the y axis. This means that both y intercepts are positive, and (2) is impossible.
So, we know that both slopes are not positive.
Supposed both slopes are negative.
the same logic shows us that if both x intercepts are positive, then both y intercepts are positive. Both x intercepts cannot be negative in this case.

So we have disproved each case in which we assume that the slopes are both positive or both negative, so one slope is positive and one is negative, and their product is definitely negative.

Answer 4

Others have answered it correctly but in GMAT we do not have that much time to think.

let the intercepts of the lines are x1 x2 and y1 y2

now x1 x2 > 0

y1 y2 < 0

we need to find product of slopes

-y1/x1 * - y2/x2 = y1y2/x1x2 <0 so product is -ve

point of intersection (4,3) is redundant information and one should not misguided
to compute the slope and other details.

Answer 5

Let line K be y = mK x + bK
and line L be y = mL x + bL

Since (4, 3) lies on both lines

4mK + bK = 3 → bK = 3 - 4mK
and 4mL + bL = 3 → bL = 3 - 4mL

Whence line K is given by: y = mK x + 3 - 4mK
ie x/(4 - 3/mK) + y/(3 - 4mK) = 1
And line L is given by y = mL x + 3 - 4mL
ie x/(4 - 3/mL) + y/(3 - 4mL) = 1

So for lineK:
x-intercept is (4 - 3/mK, 0)
y-intercept is (0, 3 - 4mK)

Similarly for line L
x-intercept is (4 - 3/mL, 0)
y-intercept is (0, 3 - 4mL)

So product of x intercepts = (4 - 3/mK)(4 - 3/mL)
= 16 - 12(1/mK + 1/mL) + 9/(mK.mL)
= 16 - [12(mK + mL) + 9)]/(mK.mL) > 0

So [12(mK + mL) + 9]/(mK.mL) < 16
ie 16(mK.mL) >12(mK + mL) + 9 if mK.mL > 0
OR 16(mK.mL) <12(mK + mL) + 9 if mK.mL < 0

Also product of y intercepts = (3 - 4mK)(3 - 4mL)
= 9 - 12(mK + mL) + 16mK.mL <0
So 16mK.mL< 12(mK + mL) - 9

So12(mK + mL) - 9>12(mK + mL) + 9 if mK.mL > 0
But this is absurd as it implies -9 > 9

So mK.mL < 0

Hence the product of their slopes negative

Answer 6

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This would be your situation if both products would have been positive. Remember that the product is positive is both are positive or both are negative. A product is negative is one of them is negative and the other one is positive

So, since one of them is negative, one of the lines must have a negative slope, their product is negative

Anabel