andresfigue wrote:

Hi, i need hel with these two problems!!!

A). If ab different from 0 and points (-a,b) and (-b,a) are in the same quadrant of the xy-plane, is point (-x,y) in the same quadrant?

1)xy>0

2)ax>0

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

1).The product of the x-intersecets of lines L and K is posstive

2). The product of the y-intersecets of lines L and K is negative

Hi, and welcome to the Gmat Club. Below are the solutions to your questions:

A. If ab different from 0 and points (-a,b) and (-b,a) are in the same quadrant of the xy-plane, is point (-x,y) in the same quadrant?The fact that points \((-a,b)\) and \((-b,a)\) are in the same quadrant means that \(a\) and \(b\) have the same sign. These points can be either in II quadrant, in case \(a\) and \(b\) are both positive, as \((-a,b)=(-,+)=(-b,a)\) OR in IV quadrant, in case they are both negative, as \((-a,b)=(+,-)=(-b,a)\) ("=" sign means here "in the same quadrant").

Now the point \((-x,y)\) will be in the same quadrant if \(x\) has the same sign as \(a\) (or which is the same with \(b\)) AND \(y\) has the same sign as \(a\) (or which is the same with \(b\)). Or in other words if all four: \(a\), \(b\), \(x\), and \(y\) have the same sign.

Note that, only knowing that \(x\) and \(y\) have the same sign won't be sufficient (meaning that \(x\) and \(y\)

must have the same sign but their sign

must also match with the sign of \(a\) and \(b\)).

(1) \(xy>0\) --> \(x\) and \(y\) have the same sign. Not sufficient.

(2) \(ax>0\) --> \(a\) and \(x\) have the same sign. But we know nothing about \(y\), hence not sufficient.

(1)+(2) \(x\) and \(y\) have the same sign AND \(a\) and \(x\) have the same sign, hence all four \(a\), \(b\), \(x\), and \(y\) have the same sign. Thus point \((-x,y)\) is in the same quadrant as points \((-a,b)\) and \((-b,a)\). Sufficient.

Answer: C.

B. 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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