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Re: Lines m and n lie in the xy-plane and intersect at the point (-2; 4). [#permalink]
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I would go with D.
We are already given one point at the point of intersection.
Statement 1 gives us an extra point therefore we can determine which line is steeper. sufficient
Statement 2 also gives us the same information. sufficient
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Re: Lines m and n lie in the xy-plane and intersect at the point (-2; 4). [#permalink]
liftoff wrote:
I would go with D.
We are already given one point at the point of intersection.
Statement 1 gives us an extra point therefore we can determine which line is steeper. sufficient
Statement 2 also gives us the same information. sufficient


I was thinking if determining which one is steeper is enough or not.

Per (1) The x-intercept of line m is greater than the x-intercept of line n.

so line m can have a slope of lets say -2. If line n also has a -ve slope then it will need to be 'flatter' than line m for its x intercept to be greater than that of line m. So its slope will need to be > -2 (for -ve slopes flatter line are closer to 0 that slopes of steeper lines). so in this case slope of line n can be something like -1. So slope of n > slope of m (-1>-2).
however, if slope of m is lets say 2, slope of n can be -ve like -2 with a greater x intercept. This satisfies the condition 1, but slope of m> slope of n in this case (2>-2).
Hence, (1) is insufficient.

Similarly we can prove for the y intercept in case of 2nd statement.

The 2 statements taken together, they should still be insufficient.

OA is D. But i'm having trouble understanding it.

Can someone please explain if I'm wrong?
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Lines m and n lie in the xy-plane and intersect at the point (-2; 4). [#permalink]
I have seen a lot of examples of this question, and every time it comes down to the Y intercept.
Can we use this as a shortcut? can we make a generalization here?
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Re: Lines m and n lie in the xy-plane and intersect at the point (-2; 4). [#permalink]
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ronr34 wrote:
Bunuel wrote:
Bumping for review and further discussion.

I have seen a lot of examples of this question, and every time it comes down to the Y intercept.
Can we use this as a shortcut? can we make a generalization here?


Yes we can use something else other than the graph method, which I find a little bit difficult to follow on many cases.

Let's see we need to know if m>n or if m-n>0, 'm and 'n' being the slopes of the respective lines

We know that they intersect at point (-2,4)

Then we have Line m= y = mx + b ---> 4 = -2m +b
Line n = y = nx+c---> 4=-2n+c

Both equal, -2n+c= -2m+b
2m-2n = b-c
m-n = (b-c)/2

Now going back to the question.

Is b-c/2 > 0?

is b-c>0, is b-c?

Statement 1

We are given that -b/m>-c/n
-bn>-cm

We can't tell whether b>c
Insuff

Statement 2

This is exactly what we were looking for
b>c

Sufficient

B stands

Hope its clear
Cheers!
J :)

YES YOU CAN

Originally posted by jlgdr on 28 Dec 2013, 16:14.
Last edited by jlgdr on 18 May 2014, 09:05, edited 1 time in total.
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Re: Lines m and n lie in the xy-plane and intersect at the point (-2; 4). [#permalink]
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amaxing Jlgdr!! thank you buddy. it is so clear now
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Re: Lines m and n lie in the xy-plane and intersect at the point (-2; 4). [#permalink]
Bunuel wrote:

(1) The x-intercept of line m is greater than the x-intercept of line n --> \(-\frac{b}{m}>-\frac{c}{n}\)


hi Bunuel, I'm not sure how can we come to this equation mentioned above. Would you please explain a bit on this ?
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Re: Lines m and n lie in the xy-plane and intersect at the point [#permalink]
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musunna wrote:
Bunuel wrote:

(1) The x-intercept of line m is greater than the x-intercept of line n --> \(-\frac{b}{m}>-\frac{c}{n}\)


hi Bunuel, I'm not sure how can we come to this equation mentioned above. Would you please explain a bit on this ?


I think this is explained here: slopes-of-m-and-n-124025.html#p1029568

Equation of a line in point intercept form is \(y=mx+b\), where: \(m\) is the slope of the line, \(b\) is the y-intercept of the line (the value of \(y\) for \(x=0\)),\(-\frac{b}{m}\) is the x-intercept of the line (the value of \(x\) for \(y=0\)). (Check Coordinate Geometry chapter of Math Book for more on this topic: math-coordinate-geometry-87652.html)
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Re: Lines m and n lie in the xy-plane and intersect at the point (-2; 4). [#permalink]
Bunuel wrote:
karthiksms wrote:
could someone pl explain ? i'm unable to understand answer.


ALGEBRAIC WAY:
Lines m and n lie in the xy-plane and intersect at the point (-2; 4). Is the slope of line m less than the slope of line n?

Equation of a line in point intercept form is \(y=mx+b\), where: \(m\) is the slope of the line, \(b\) is the y-intercept of the line (the value of \(y\) for \(x=0\)), \(-\frac{b}{m}\) is the x-intercept of the line (the value of \(x\) for \(y=0\)). (Check Coordinate Geometry chapter of Math Book for more on this topic: https://gmatclub.com/forum/math-coordina ... 87652.html)

We are given two lines: \(y_m=mx+b\) and \(y_n=nx+c\). Now, as they intersect at the point (-2; 4) then: \(4=-2m+b\) and \(4=-2n+c\) (this point is common for both of the lines) --> \(b=4+2m\) and \(c=4+2n\).

Question: is \(m<n\)?

(1) The x-intercept of line m is greater than the x-intercept of line n --> \(-\frac{b}{m}>-\frac{c}{n}\) --> \(-\frac{4+2m}{m}>-\frac{4+2n}{n}\) --> \(\frac{1}{n}-\frac{1}{m}>0\) --> insufficient to answer whether \(m<n\): if \(n=2\) and \(m=-4\) then YES but if \(n=2\) and \(m=4\) then NO. Not sufficient.

(2) The y-intercept of line n is greater than the y-intercept of line m --> \(c>b\) --> \(4+2n>4+2m\) --> \(n>m\). Sufficient.

Answer: B.


why can't we deduce like below -
1/n-1/m >0=> 1/n>1/m=> m>n

Can you please explain.
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Re: Lines m and n lie in the xy-plane and intersect at the point (-2; 4). [#permalink]
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02999 wrote:
Bunuel wrote:
karthiksms wrote:
could someone pl explain ? i'm unable to understand answer.


ALGEBRAIC WAY:
Lines m and n lie in the xy-plane and intersect at the point (-2; 4). Is the slope of line m less than the slope of line n?

Equation of a line in point intercept form is \(y=mx+b\), where: \(m\) is the slope of the line, \(b\) is the y-intercept of the line (the value of \(y\) for \(x=0\)), \(-\frac{b}{m}\) is the x-intercept of the line (the value of \(x\) for \(y=0\)). (Check Coordinate Geometry chapter of Math Book for more on this topic: https://gmatclub.com/forum/math-coordina ... 87652.html)

We are given two lines: \(y_m=mx+b\) and \(y_n=nx+c\). Now, as they intersect at the point (-2; 4) then: \(4=-2m+b\) and \(4=-2n+c\) (this point is common for both of the lines) --> \(b=4+2m\) and \(c=4+2n\).

Question: is \(m<n\)?

(1) The x-intercept of line m is greater than the x-intercept of line n --> \(-\frac{b}{m}>-\frac{c}{n}\) --> \(-\frac{4+2m}{m}>-\frac{4+2n}{n}\) --> \(\frac{1}{n}-\frac{1}{m}>0\) --> insufficient to answer whether \(m<n\): if \(n=2\) and \(m=-4\) then YES but if \(n=2\) and \(m=4\) then NO. Not sufficient.

(2) The y-intercept of line n is greater than the y-intercept of line m --> \(c>b\) --> \(4+2n>4+2m\) --> \(n>m\). Sufficient.

Answer: B.


why can't we deduce like below -
1/n-1/m >0=> 1/n>1/m=> m>n

Can you please explain.


You cannot cross-multiply 1/n > 1/m to get m > n because we don't know the signs of m and n. If they have the same sign, then yes, after cross-multiplying we get m > n but if n is positive and m is negative, then after cross-multiplying we get m < n. For example:

If n = 2 and m = 3 --> 1/2 > 1/3 --> 3 > 2;
If n = -3 and m = -2 -->1/(-3) > 1/(-2) --> -2 > -3.

But if n = 2 and m = -2 -->1/2 > 1/(-2) --> -2 < 2.
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Re: Lines m and n lie in the xy-plane and intersect at the point (-2; 4). [#permalink]
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