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If l and K are lines in xy- plane, is the product of the slope of l and k equal to -1?

1. Line l passes through the origin and the point (1,2). 2. Line K has x-intercept 4 and y-intercept 2.

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.

If l and K are lines in xy- plane, is the product of the slope of l and k equal to -1?

1. Line l passes through the origin and the point (1,2). 2. Line K has x-intercept 4 and y-intercept 2.

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.

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.

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

We don't need to calculate anything to answer this question.

If L and K are lines in xy- plane, is the product of the slope of l and k equal to -1?

The product of two lines equal to -1 if these lines are perpendicular to each other. So, the question basically asks whether L and K are perpendicular to each other.

Now, if we knew the equations of each line we would be able to answer the question. What do we need to find equation of a line? Since a line is defined by two points, then if we knew any two points of a line then we could get its equatiom

(1) Line L passes through the origin and the point (1,2). We have two points of line L: (0, 0) and (1, 2). So, we can get the equation of line L, though we know nothing about line K. Not sufficient.

(2) Line K has x-intercept 4 and y-intercept 2. We have two points of line K: (4, 0) and (0, 2). So, we can get the equation of line K, though we know nothing about line L. Not sufficient.

(1)+(2) We know both equation, hence we can answer whether the lines are perpendicular to each other. Sufficient.

Answer: C.

venmic wrote:

Can anyone give a better explanation on this one please

Hovv does K have a slope of -1/2 out of novvhere at all

Please any expert

Given two points \((x_1,y_1)\) and \((x_2,y_2)\) on a line, the slope \(m\) of the line is:

\(m=\frac{y_2-y_1}{x_2-x_1}\).

Since x-intercept of line K is 4, then it passes through point (4, 0) and since y-intercept of line K is 2, the it passes through point (0, 2). Hence it's slope is \(m=\frac{2-0}{0-4}=-\frac{1}{2}\).

Re: If L and K are lines in xy- plane, is the product of the [#permalink]

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20 Oct 2012, 14:46

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.

Re: If L and K are lines in xy- plane, is the product of the [#permalink]

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27 Jul 2015, 13:37

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Re: If L and K are lines in xy- plane, is the product of the [#permalink]

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28 Nov 2016, 21:56

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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