Well....its twice the horror for me man....below is the best I could come up with. If anyone can explain the solution clearly would be great. If this comes in the real thing, I'll most likely guess instead of trying to solve it.
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Slope of line = (y2-y1)/(x2-x1)
We can assume (x2, y2) as (4,3)

From (1) either both x-intercepts are +ve or -ve so either (x1, 0) and (x2, 0)
If +ve, slope of L = (3 - 0)/(4 - x1) ... depends on value of x1
If -ve, slope of L = (3 - 0)/(4 - x1) ... which is positive.

Same goes for Line K.
Hence insufficient.

From (2) either one y-intercept is +ve and another is -ve so (0, y1) and (0, -y2)

Slope of L = (3 - y1)/(4 - 0) ... depends on value of y1
Slope of K = (3 - (-y2))/(4 - 0) ... which is positive

Hence insufficient

Taking both together, we still cannot say due to values of x1 and y1 unknown hence answer should be E.

E should be the answer.

Deduced with help of rough sketch. Tried to make one with mspaint but too much hard work

Just thought of taking diagramatic approach for this & getting "C". Looks like following is the only possible solution.

The question becomes much simpler if you see whether the lines passing through point (4, 3) has any bearing on the sign of the slopes -- NO.

So, it comes down to the simple slope equation:
slope = y-intercept / x-intercept.

We need slope(l)*slope(k)

slope(l)*slope(k) = ((y-intercept of l) / x-intercept of l) * ((y-intercept of k / x-intercept of k)
= (y-intercept of l * y-intercept of k) / (x-intercept of l * y-intercept of x)

Using 1) and 2),
= (-ve) (+ve)
= -ve

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

We have two lines: y_l=m_1x+b_1 and y_k=m_2x+b_2. The question: is m_1*m_2<0?

Lines intersect at the point (4,3) --> 3=4m_1+b_1 and 3=4m_2+b_2

(1) The product of the x-intersects of lines L and K is positive. Now, one of the lines can intersect x-axis at 0<x<4 (positive slope) and another also at 0<x<4 (positive slope), so product of slopes also will be positive BUT it's also possible one line to intersect x-axis at 0<x<4 (positive slope) and another at x>4 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: x-intersect is value of x for y=0 and equals to x=-\frac{b}{m} --> so (-\frac{b_1}{m_1})*(-\frac{b_2}{m_2})>0 --> \frac{b_1b_2}{m_1m_2}>0.

(2) The product of the y-intersects of lines L and K is negative. Now, one of the lines can intersect y-axis at 0<y<3 (positive slope) and another at y<0 (positive slope), so product of slopes will also be positive BUT it's also possible one line to intersect y-axis at y<0 (positive slope) and another at y>3 (negative slope) and in this case product of slopes will be negative. Two different answers, hence not sufficient.

But from this statement we can deduce the following: y-intercept is value of y for x=0 and equals to x=b --> b_1*b_2<0.

(1)+(2) \frac{b_1b_2}{m_1m_2}>0 and b_1*b_2<0. As numerator in \frac{b_1b_2}{m_1m_2}>0 is negative, then denominator m_1m_2 must also be negative. So m_1m_2<0. Sufficient.

Answer: C.

In fact we arrived to the answer C, without using the info about the intersection point of the lines. So this info is not needed to get C.

For more on coordinate geometry check the link in my signature.
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X-intercept is the point where a line (a graph) crosses the x-axis. So it's the point on x-axis, any point on x-axis has y-coordinate equal to zero, which means that X-intercept is the point (x,0) - the value of x when y=0: y=mx+b --> 0=mx+b --> x=-\frac{b}{m}. So X-intercept of a line y=mx+b is x=-\frac{b}{m};

Y-intercept is the point where a line (a graph) crosses the y-axis. So it's the point on y-axis, any point on y-axis has x-coordinate equal to zero, which means that Y-intercept is the point (0,y) - the value of y when x=0: y=mx+b --> y=m*0+b --> y=b. So Y-intercept of a line y=mx+b is y=b.

Check Coordinate Geometry chapter for more (link in my signature).

Hope it's clear.
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory

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DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!


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whats the difficulty level of this question???? Are such type of questions asked in actual gmat??Mans are these above 50 points level qustions??

Question: Product of the slope of line l and k -ve.

Let line l be y1= m1x1+c1 and k be y2 = m2x2+c2. General equation of the line => y = mx+c.

Hence to determine whether m1*m2 = -ve

Now x intercept is when y=0 and y intercept is when x=0.

Statement A: Product of x intercepts of line l and k is +ve.

x-intercept of line l, that is when y=0=> x1=-c1/m1 and x-intercept of line k is x2=-c2/m2.

Given their product is +ve. Now -c1/m1 * -c2/m2 is +ve. Either c1*c2 and m1*m2 could be both +ve or both -ve. We cannot determine the whether m1*m2 is +ve or -ve since we do not have information about c1*c2.

Statement B: Product of y intercepts of line l and k is -ve.

y-intercept of line l, that is when x=0=> y1=c1 and y-intercept of line k is y2=c2.

Given their product is -ve that is c1*c2=-ve. However with this statement alone we cannot determine whether m1*m2=-ve or +ve.

Combining both the statement we know that since c1*c2 = -ve, m1*m2 has to be -ve to satisfy Statement A.

Hence answer is C.
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Equation for a line in intercept form, X/A+Y/B=1 and slope will be m=-(B/A)

Now take the equation of first line as X/A1+Y/B1=1 and second line as X/A2+Y/B2=1

Product of their slope will be (B1/A1)*(B2/A2)=(B1*B2)/(A1*A2)

with first (A1 *A2) positive
With 2nd (B1*B2) -ve

So with both we can say Product of their slope is -ve. that is C.

Consider KUDOS if you like my solution.