If l and k are lines in the xy-plane, is the product of the slopes of

C


1: we dont know anything about the other line but we can find the slope of line l = 2

2: again we dont know the other line's slope but k's is -1/2

Together 2*-1/2= -1

But you could just realize that 1 and 2 are suff together b/c we know the slopes we are going to get a YES OR A NO NOT BOTH. so u could save some time.

(4,0) and (0,2) --> (0-2)/(4-0) --> -2/4 --> -1/2

Formula for slope is (y1-y2)/(x1-x2)

>> Line K has x-intercept 4 and y-intercept 2.

Its not clear if the x-intercept is going to be a (4,0). All it said is that x-intercept is 4 and y-intercept is 2. Even a coordinate of (-4,0) equals an x-intercept of 4.



So the x coordinates can be (4,0) or (-4,0) and y coordinates can be (0,2) or (0,-2)

So there are two possible lines that are perpendicular and the other 2 are not.

So in my opinion E.

What is the OA ?

my mistake if 1) is true then 2) will have a specific set of coordinates.

C is correct...

Need to be more cautious.

what do u mean by 2 has to have a specific set of cordinates .. form what i see i agree to your earlier assumption ...when u say line k has 4 as x cordinate and 2 as y cordinate u can very well say that it is a cordinate of the same point (4,2) then u cannot say anything about K's slope as it just passes through one point .. please explain as i still feel it should be E

read 2) carefully - line K has an x-intercept of 4 (meaning it hits the x axis at 4--- so the point would therefore be 4,0) and y-intercept of 2 (meaning it hits the y axis at 2---so the point would therefore be 0,2). So we now have (4,0) and (0,2) as our two points for line K and we can plug it into our slope formula of y2 - y1/x2 - x1 --> y2 = 2 y1 = 0 x2 - 0 y1 = 4 --> 2-0/0-4 ---> 2/-4 = -1/2.


Hope this helps!

for this type of question, we do not use algebra until we can not solve the problem by using logic reasoning
from both choice we see that both lines are determined , this mean we know the slope of each lines and can know whether product of the lines are negative.

Asked: If l and k are lines in the xy-plane, is the product of the slopes of l and k equal to -1 ?

(1) Line l passes through the origin and the point (1,2).
Equation of line l : y = 2x
But line k is unknown
NOT SUFFICIENT

(2) Line k has x-intercept 4 and y-intercept 2.
Equation of line k : x/4 + y/2 = 1
But line l is unknown
NOT SUFFICIENT

(1) + (2)
(1) Line l passes through the origin and the point (1,2).
Equation of line l : y = 2x
(2) Line k has x-intercept 4 and y-intercept 2.
Equation of line k : x/4 + y/2 = 1 or y = -x/2 + 2
Product of the slopes of l and k = 2* (-1/2) = - 1
SUFFICIENT

IMO C

Hi BrentGMATPrepNow, regarding line K of (4,0) (0,2), it seems the line can be drawn either vertically or horizontally on point of (4,2). Therefore it doesn't seem line K is perpendicular to line L at the crossing point unless is at the origin point? Have I missed something in the reasoning here?

Also not quite sure with question asked is the product of the slope of L and K equal to -1? Since we got slope = -1/2. Where or how is the product here? Thanks Brent

Are you referring to the fact that line K has x-intercept 4 and y-intercept 2 (ie, statement 2)?
If so, then statement to is telling us that line K passes through BOTH of the points (4,0) and (0,2), and not just one of them.

Thanks BrentGMATPrepNow. My bad forgot to mention statement 2. Yes and understand now.
By the way what about the product of the slope of L and K equal to -1? Since we got slope = -1/2. Where or how is the product here? Thanks Brent

From statement 1, we can conclude that the slope of line L is 2.
From statement 2, we can conclude that the slope of line K is -1/2.

The target question asks "Is the product of the slope of L and K equal to -1?"
In other words, does (2)(-1/2) = -1?
The answer is YES, which means the combined statements are sufficient.

Question Stem Analysis:

We need to determine whether the product of the slopes of lines l and k is equal to -1. Remember that if the product of the slopes of two lines is -1, then those two lines are perpendicular.

Statement One Alone:

\(\Rightarrow\) Line l passes through the origin and the point (1,2).

Since we know two points belonging to line l, we can calculate the slope of line l. However, since we know nothing about the slope of line k, this is not sufficient to answer the question.

Eliminate answer choices A and D.

Statement Two Alone:

\(\Rightarrow\) Line k has x-intercept 4 and y-intercept 2.

Since we know two points belonging to line k, we can calculate the slope of line k. However, since we know nothing about the slope of line l, this is not sufficient to answer the question.

Eliminate answer choice B.

Statements One and Two Together:

Using statement one, we can calculate the slope of line l, and using statement two, we can calculate the slope of line k. Since we can calculate the slopes of both lines, we can answer the question of whether the product of the two slopes is equal to -1.

Remember that in DS questions, we don't have to answer the question, we only need to determine whether we have enough information to do so. It would be a huge waste of time to actually calculate the slopes of the two lines in this question when it is clear that neither statement is sufficient on its own, but both statements together are sufficient.

Answer: C

Generic question Bunuel KarishmaB avigutman : Per my understanding, on the X – Y graph, every line on the X- Y axis has the following template :

Y = mx + constant

But for this line on the graph below , the equation of the line is x = 5

Question : how to go from y = mx + constant to x =5 ?

Is Y = 0 ?
Is m = (any number) / 0 ?
Is Constant = 0 ?

If Yes, then I plug that into the template equation :

Y (=0) = M (any number/0). x + constant (=0)

or

0 = (any number/0) x + 0



I dont see how you get x = 5 from y = mx + constant.

I am struggling to go from y = mx + constant to x = 5

Think of it in the reverse way. A set of points that satisfy a certain condition is called the locus of the condition.
The locus of y = mx + c is a straight line where m and c are real numbers.
The locus on x = a is also a straight line parallel to y axis.
Since m is not defined (infinity) in this case, we do not represent it in y = mx + c form.

Great explanation BrentGMATPrepNow and crystal clear now. Thanks Brent

See page 220 in my book, jabhatta2.