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Question asks if two lines perpendicular. Yes/No

St1. x intercepts of line l and k are positive. Lines may locate on many options, including perpendicular manner. Answer Yes and No. So, INSUFFICIENT

St.2 y intercept of line l and k are negative. Both lines cannot have negative y interception if they are perpendicular. Answer No. SUFFICIENT

B
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line 1:
\(y = m_1*x + b_1\)
\(4 = 3m_1 + b_1\)

line 2:
\(y = m_2*x + b_2\)
\(4 = 3m_2 + b_2\)

Therefore the question is reduced to is \(m_1 * m_2 = -1\)?

Statement 1) x intercepts >0. The x-intercept is when y = 0, so x intercept is \(\frac{-m_1}{b_1}\). Therefore \(\frac{-m_1}{b_1}> 0\) and \(\frac{-m_2}{b_2}> 0\) Therefore we know that the slope and y-intercept must have opposite signs, but the slopes could still be anything. Insufficient.

Statement 2) y intercepts <0. \(b_1\) and \(b_2\) are both negative. Let's pretend it's the smallest negative number (very close to zero). Then 4\(= 3m_1 + b_1 < 3m_1\), so \(4<3m_1\), and \(m_1 > \frac{4}{3}\) Same with \(m_2\)

This tells us that the slopes of both lines are positive and greater than 4/3. Therefore their product cannot possibly be -1. Sufficient.

Answer: B
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We know the lines pass through a certain point. To be able to conclude whether their respective slopes multiplied by each other equals -1 we would need to know the slopes or how they relate to one another.

This question makes it seem as if we need to find the slopes and multiply them, however, we can answer the question without knowing the slopes or even their final product. This is because we can make general inferences that exclude and include certain values to be possible.

If the slope product is -1, then the lines have reciprocal slopes with different signs. This only happens when a line is perpendicular to another line.

A) tells us nothing about the slope of either line. Passing through 4,3 the lines can still have positive x-intercepts with a range of different slopes. The product is not possible to find given the circumstances.

B) tells us that the y-intercept is negative. This would require both lines to travel upwards in order to get to 4,3. Thus, both slopes are positive and even though we cannot give the actual slope of either line, we can conclude that +*+ will never be negative.

Answer choice B.
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PathFinder007
Two line l and k intersect at a point (4, 3). Is the product of their slopes -1?

(1) x intercepts of line l and k are positive
(2) y intercept of line l and k are negative

This one is tricky... nice!

\(\left( {4,3} \right)\,\, \in \,\,\,\left( {{\text{lin}}{{\text{e}}_{\,l}}\,\, \cap \,\,\,{\text{lin}}{{\text{e}}_{\,k}}} \right)\)

\({\text{slop}}{{\text{e}}_{\,l}}\,\, \cdot \,\,\,{\text{slop}}{{\text{e}}_{\,k}}\,\,\,\mathop = \limits^? \,\,\, - 1\)

1) Insufficient. We present a GEOMETRIC BIFURCATION in the image attached.

(2) We know line l and line k must have positive slopes (!), therefore the product of their slopes is positive (and not equal to -1)!

The right answer is therefore (B).


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Attachments

13Set18_9m.gif
13Set18_9m.gif [ 20.59 KiB | Viewed 7549 times ]

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Two line l and k intersect at a point (4, 3). Is the product of their slopes -1?

(1) x intercepts of line l and k are positive
(2) y intercept of line l and k are negative


I'm not sure about the source of this question.
Finding it hard to visualize while trying to solve it even after referring to the following similar(IMO) question-
https://gmatclub.com/forum/lines-n-and- ... 27999.html

Any help is appreciated.
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It's better to do this question by visualizing all the scenarios. We can solve it algebraically tho. :)

Equation of line l
y=Lx+3-4L

Equation of line k
y=Kx+3-4K

1) x intercept of line l= \(\frac{4L-3}{L}\)
\(\frac{4L-3}{L}>0\)
There are 2 cases possible
1. L>0 and 4L-3>0 or L>3/4........ L>3/4
2. L<0 and 4L-3<0 or L<3/4.........L<0

Similarly we can find 2 cases for line K, when x-intercept of Line K is positive
1. K>3/4
2. K<0

When L=1 and K=-1, LK=-1
When L=1 and K=1, LK=1

Insufficient

2) 3-4L<0
L>3/4

Also, 3-4K<0
K>3/4

Both L and K are positive; hence, their product can never be negative or -1

Sufficient





geezus24x7
Two line l and k intersect at a point (4, 3). Is the product of their slopes -1?

(1) x intercepts of line l and k are positive
(2) y intercept of line l and k are negative


I'm not sure about the source of this question.
Finding it hard to visualize while trying to solve it even after referring to the following similar(IMO) question-
https://gmatclub.com/forum/lines-n-and- ... 27999.html

Any help is appreciated.
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PathFinder007
Two line l and k intersect at a point (4, 3). Is the product of their slopes -1?

(1) x intercepts of line l and k are positive
(2) y intercept of line l and k are negative


Given: Two line l and k intersect at a point (4, 3).

Asked: Is the product of their slopes -1?

If the lines intersect at a point (4,3)
y-3 = m (x -4) = mx - 4m
y = mx + (3-4m)

Let line l be y = m1x + (3 - 4m1)
Let line k be y = m2x + (3 - 4m2)


(1) x intercepts of line l and k are positive
For x-intercepts, y = 0
m1x1 + (3 - 4m1) = 0
x1 = (4m1 -3) / m1 = 4 - 3/m1 > 0
3/m1 < 4; m1<>0; Either m1<0 or m1 >3/4
3/m2 < 4; m2<>0; Either m2<0 or m2 >3/4
m1 * m2 may or may not be -1
NOT SUFFICIENT

(2) y intercept of line l and k are negative
For y intercept, x=0
y1 = 3 - 4m1 < 0; m1>3/4
y2 = 3 - 4m2 < 0; m2>3/4
m1 * m2 >9/16 and is NOT = -1
SUFFICIENT

IMO B
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PathFinder007
Two line l and k intersect at a point (4, 3). Is the product of their slopes -1?

(1) x intercepts of line l and k are positive
(2) y intercept of line l and k are negative

solve via plotting the given info
#2 would be sufficient
IMO B
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