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

If two lines are perpendicular to each other, the slope of one will be the negative reciprocal of the slope of the other (that is, if one slope is m, the other slope will be -1/m). 'Is the product of their slopes -1?' is just another way of asking whether this is true, since m(-1/m) = -1. So, translate the question as "are the two lines perpendicular?"

Now, let's draw. The goal is to prove insufficiency, so we'll be trying to draw, if possible, two pairs of lines for each statement: one pair that fits the statement and IS perpendicular, and one pair that fits the statement and ISN'T perpendicular. If that's impossible, then the statement must be sufficient.

Statement 1: both x intercepts are positive. That is, they both intercept the x axis to the right of 0.

Here's a perpendicular pair of lines that fits:



Here's a non perpendicular pair of lines that fits:



Statement 2: Both y intercepts are negative. That is, they both intercept the y axis below 0.

Here's a NON perpendicular pair of lines that fits:



However, it isn't possible to draw a pair of perpendicular lines that fit both the info in the question stem (intersect at (4,3)) and this statement. The angle between the two lines, marked here, is always going to be significantly less than 90 degrees if we only use lines that have negative y intercepts.



Therefore, using this statement, the answer to the question is definitely "no." So, the statement is sufficient and the answer is B.
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Hi PathFinder007,

When dealing with graphing questions on the GMAT, it often helps to draw sketches/pictures to help you stay organized. If you're comfortable with graphing "rules" though, then you don't necessarily need a drawing.

Here, we're told that two lines intercept at the point (4,3). Since we don't know anything about the individual lines (K and L), these lines could have positive slopes, negative slopes, 0 slopes (meaning the lines are horizontal) or undefined slopes (meaning the lines go "straight up and down"). We're asked if the product of their slopes is -1. This is a YES/NO question.

There's a great opportunity here to "rewrite" the question. The ONLY time that the slopes of 2 lines produce a product of -1 is when the two lines are PERPENDICULAR. So this question is essentially asking if lines K and L are perpendicular to one another (meaning they cross and form a 90 degree angle). Perpendicular lines have slopes that can be called "opposite inverse" or "negative reciprocal" (for example, slopes of -2 and +1/2).

Fact 1: The X-intercepts of both lines are positive.

This tells us very little about the two lines. They could have positive or negative slopes (or even an undefined slope).

IF....
X-intercept of K = 1
X-intercept of L = 2
The lines are NOT perpendicular and the answer to the question is NO.

IF.....
X-intercept of K = 1
X-intercept of L = 7
The lines ARE perpendicular and the answer to the question is YES.
Fact 1 is INSUFFICIENT

Fact 2: The Y-intercepts of both lines are negative.

This Fact proves that the slopes of BOTH lines are positive (since they cross the Y-axis at a negative number then move UP and to the RIGHT to the point (4,3)), so there is NO WAY for them to cross and be perpendicular. The answer to the question is ALWAYS NO.
Fact 2 is SUFFICIENT.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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

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

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.

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.

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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.

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

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

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

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