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Re: Lines n and p lie in the xy-plane. Is the slope of line n [#permalink]
23 Feb 2012, 07:11
12
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BANON wrote:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p ?
(1) Lines n and p intersect at the point (5 , 1). (2) The y-intercept of line n is greater than the y-intercept of line p.
Algebraic approach:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p?
We have two lines: \(y_n=m_1x+b_1\) and \(y_p=m_2x+b_2\). Q: \(m_1<m_2\) true?
(1) Lines n and p intersect at the point (5,1) --> \(1=5m_1+b_1=5m_2+b_2\) --> \(5(m_1-m_2)=b_2-b_1\). Not sufficient.
(2) The y-intercept of line \(n\) is greater than the y-intercept of line \(p\) --> y-intercept is value of \(y\) for \(x=0\), so it's the value of \(b\) --> \(b_1>b_2\) or \(b_2-b_1<0\). Not sufficient.
(1)+(2) \(5(m_1-m_2)=b_2-b_1\), as from (2) \(b_2-b_1<0\) (RHS), then LHS (left hand side) also is less than zero \(5(m_1-m_2)<0\) --> \(m_1-m_2<0\) --> \(m_1<m_2\). Sufficient.
Re: Lines n and p lie in the xy-plane. Is the slope of line n [#permalink]
23 Feb 2012, 07:13
4
This post received KUDOS
Expert's post
3
This post was BOOKMARKED
BANON wrote:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p ?
(1) Lines n and p intersect at the point (5 , 1). (2) The y-intercept of line n is greater than the y-intercept of line p.
Graphic approach:
Lines n and p lie in the xy plane. Is the slope of line n less than the slope of line p?
(1) Lines n and p intersect at (5,1) (2) The y-intercept of line n is greater than y-intercept of line p
The two statements individually are not sufficient.
(1)+(2) Note that a higher absolute value of a slope indicates a steeper incline.
Now, if both lines have positive slopes then as the y-intercept of line n (blue) is greater than y-intercept of line p (red) then the line p is steeper hence its slope is greater than the slope of the line n:
Attachment:
1.PNG [ 14.29 KiB | Viewed 11743 times ]
If both lines have negative slopes then again as the y-intercept of line n (blue) is greater than y-intercept of line p (red) then the line n is steeper hence the absolute value of its slope is greater than the absolute value of the slope of the line p, so the slope of n is more negative than the slope of p, which means that the slope of p is greater than the slope of n:
Attachment:
2.PNG [ 13.66 KiB | Viewed 11724 times ]
So in both cases the slope of p is greater than the slope of n. Sufficient.
Re: Lines n and p lie in the xy-plane. Is the slope of line n [#permalink]
23 Feb 2012, 09:13
1
This post received KUDOS
Expert's post
nglekel wrote:
Bunuel,
What if line p has a negative y intercept but line n has a positive intercept? Wouldn't that give the oposite answer?
If line p has a negative y-intercept then its slope is positive and it will still be more than the slope of n, with positive y-intercept (if the slope of n will be positive than p will still be steeper than n, and if the slope of n is negative it obviously will be less than positive slope of p). Consider first image and rotate line n (blue) so that it to have positive y-intercept and you'll easily see the answer.
Re: Lines n and p lie in the xy-plane. Is the slope of line n le [#permalink]
13 Sep 2012, 11:19
monikaleoster wrote:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p ? (1) Lines n and p intersect at the point (5,1). (2) The y-intercept of line n is greater than the y-intercept of line p.
Here we have two lines and two slopes So lets first write the equations for our lovely lines
Y = mnX + Cn Y = mpX + Cp
Now statement one says that it intersects at 5,1. SO lets put it in the equations and subtract them
We get, (mn-mp)5 = Cp-Cn that tells us nothing about the slopes of the lines or their relative values, but if we know the value of Cp-Cn that weather it is positive or negative we will know weathet mn-mp is positive or negative and that which is greater
Statement 2 Y intercept of line n is greater than p so that gives us Cn >Cp
Alone this statement is also not sufficient. it talks abou c not slopes
But if we combine the two, Voila !! we know which slope is greater.
Re: Lines n and p lie in the xy-plane. Is the slope of line n [#permalink]
29 Dec 2012, 08:18
I don't get it :/
what if slope of line P is positive and the slope of line N negative (but still satisfying all the condition...) Bunuel, on your examples the slopes have the same sign... are we talking about the absolute value of the slope?
Re: Lines n and p lie in the xy-plane. Is the slope of line n [#permalink]
16 Jul 2013, 18:23
Bunuel wrote:
BANON wrote:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p ?
(1) Lines n and p intersect at the point (5 , 1). (2) The y-intercept of line n is greater than the y-intercept of line p.
Algebraic approach:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p?
We have two lines: \(y_n=m_1x+b_1\) and \(y_p=m_2x+b_2\). Q: \(m_1<m_2\) true?
(1) Lines n and p intersect at the point (5,1) --> \(1=5m_1+b_1=5m_2+b_2\) --> \(5(m_1-m_2)=b_2-b_1\). Not sufficient. (2) The y-intercept of line \(n\) is greater than the y-intercept of line \(p\) --> y-intercept is value of \(y\) for \(x=0\), so it's the value of \(b\) --> \(b_1>b_2\) or \(b_2-b_1<0\). Not sufficient.
(1)+(2) \(5(m_1-m_2)=b_2-b_1\), as from (2) \(b_2-b_1<0\) (RHS), then LHS (left hand side) also is less than zero \(5(m_1-m_2)<0\) --> \(m_1-m_2<0\) --> \(m_1<m_2\). Sufficient.
Answer: C.
Hope it helps.
Bunuel,
In here - \(y_n=m_1x+b_1\) and \(y_p=m_2x+b_2\). Q: \(m_1<m_2\) true?
Why have you chosen different variables for the y?
Shouldnt the two equations be y=m1x+b1 and y=m2x+b2? We always form the equation from the basic form of y=mx+c wherein we substitute the values of m and c. And if that is the case, we can get the answer from statement II only.
I know I am missing something but I am not clear as to why you have picked different variables for y but not for x.
Re: Lines n and p lie in the xy-plane. Is the slope of line n [#permalink]
16 Jul 2013, 22:08
Expert's post
keenys wrote:
Bunuel wrote:
BANON wrote:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p ?
(1) Lines n and p intersect at the point (5 , 1). (2) The y-intercept of line n is greater than the y-intercept of line p.
Algebraic approach:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p?
We have two lines: \(y_n=m_1x+b_1\) and \(y_p=m_2x+b_2\). Q: \(m_1<m_2\) true?
(1) Lines n and p intersect at the point (5,1) --> \(1=5m_1+b_1=5m_2+b_2\) --> \(5(m_1-m_2)=b_2-b_1\). Not sufficient. (2) The y-intercept of line \(n\) is greater than the y-intercept of line \(p\) --> y-intercept is value of \(y\) for \(x=0\), so it's the value of \(b\) --> \(b_1>b_2\) or \(b_2-b_1<0\). Not sufficient.
(1)+(2) \(5(m_1-m_2)=b_2-b_1\), as from (2) \(b_2-b_1<0\) (RHS), then LHS (left hand side) also is less than zero \(5(m_1-m_2)<0\) --> \(m_1-m_2<0\) --> \(m_1<m_2\). Sufficient.
Answer: C.
Hope it helps.
Bunuel,
In here - \(y_n=m_1x+b_1\) and \(y_p=m_2x+b_2\). Q: \(m_1<m_2\) true?
Why have you chosen different variables for the y?
Shouldnt the two equations be y=m1x+b1 and y=m2x+b2? We always form the equation from the basic form of y=mx+c wherein we substitute the values of m and c. And if that is the case, we can get the answer from statement II only.
I know I am missing something but I am not clear as to why you have picked different variables for y but not for x.
n and p are subscripts of y's, not variables.
\(y=m_1x+b_1\) is equation of line n. \(y=m_2x+b_2\) is equation of line p.
I used subscripts simply to distinguish one equation from another. _________________
Re: Lines n and p lie in the xy-plane. Is the slope of line n [#permalink]
17 Jul 2013, 04:02
Bunuel wrote:
keenys wrote:
Bunuel wrote:
Algebraic approach:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p?
We have two lines: \(y_n=m_1x+b_1\) and \(y_p=m_2x+b_2\). Q: \(m_1<m_2\) true?
(1) Lines n and p intersect at the point (5,1) --> \(1=5m_1+b_1=5m_2+b_2\) --> \(5(m_1-m_2)=b_2-b_1\). Not sufficient. (2) The y-intercept of line \(n\) is greater than the y-intercept of line \(p\) --> y-intercept is value of \(y\) for \(x=0\), so it's the value of \(b\) --> \(b_1>b_2\) or \(b_2-b_1<0\). Not sufficient.
(1)+(2) \(5(m_1-m_2)=b_2-b_1\), as from (2) \(b_2-b_1<0\) (RHS), then LHS (left hand side) also is less than zero \(5(m_1-m_2)<0\) --> \(m_1-m_2<0\) --> \(m_1<m_2\). Sufficient.
Answer: C.
Hope it helps.
Bunuel,
In here - \(y_n=m_1x+b_1\) and \(y_p=m_2x+b_2\). Q: \(m_1<m_2\) true?
Why have you chosen different variables for the y?
Shouldnt the two equations be y=m1x+b1 and y=m2x+b2? We always form the equation from the basic form of y=mx+c wherein we substitute the values of m and c. And if that is the case, we can get the answer from statement II only.
I know I am missing something but I am not clear as to why you have picked different variables for y but not for x.
n and p are subscripts of y's, not variables.
\(y=m_1x+b_1\) is equation of line n. \(y=m_2x+b_2\) is equation of line p.
I used subscripts simply to distinguish one equation from another.
If that is the case then, from the above equations
we get b1=y-m1x and b2=y-m2x
Now from statement 2 we know that b1>b2...
therefore, y-m1x >y-m2x
which gives (m1-m2)x>0
So it can be proved from statement 2 only that m1>m2
Re: Lines n and p lie in the xy-plane. Is the slope of line n [#permalink]
17 Jul 2013, 05:07
Expert's post
keenys wrote:
Bunuel wrote:
keenys wrote:
Bunuel,
In here - \(y_n=m_1x+b_1\) and \(y_p=m_2x+b_2\). Q: \(m_1<m_2\) true?
Why have you chosen different variables for the y?
Shouldnt the two equations be y=m1x+b1 and y=m2x+b2? We always form the equation from the basic form of y=mx+c wherein we substitute the values of m and c. And if that is the case, we can get the answer from statement II only.
I know I am missing something but I am not clear as to why you have picked different variables for y but not for x.
n and p are subscripts of y's, not variables.
\(y=m_1x+b_1\) is equation of line n. \(y=m_2x+b_2\) is equation of line p.
I used subscripts simply to distinguish one equation from another.
If that is the case then, from the above equations
we get b1=y-m1x and b2=y-m2x
Now from statement 2 we know that b1>b2...
therefore, y-m1x >y-m2x
which gives (m1-m2)x>0
So it can be proved from statement 2 only that m1>m2
Where am I going wrong?
The y-intercept is the value of \(y\) for \(x=0\). You should substitute x=0 into both equations.
So, the y-intercept of line n is b1 and the y-intercept of line p is b2, from (2) we only have that b1>b2. _________________
Re: Lines n and p lie in the xy-plane. Is the slope of line n [#permalink]
28 Dec 2013, 04:57
Bunuel wrote:
BANON wrote:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p ?
(1) Lines n and p intersect at the point (5 , 1). (2) The y-intercept of line n is greater than the y-intercept of line p.
Graphic approach:
Lines n and p lie in the xy plane. Is the slope of line n less than the slope of line p?
(1) Lines n and p intersect at (5,1) (2) The y-intercept of line n is greater than y-intercept of line p
The two statements individually are not sufficient.
(1)+(2) Note that a higher absolute value of a slope indicates a steeper incline.
Now, if both lines have positive slopes then as the y-intercept of line n (blue) is greater than y-intercept of line p (red) then the line p is steeper hence its slope is greater than the slope of the line n:
Attachment:
1.PNG
If both lines have negative slopes then again as the y-intercept of line n (blue) is greater than y-intercept of line p (red) then the line n is steeper hence the absolute value of its slope is greater than the absolute value of the slope of the line p, so the slope of n is more negative than the slope of p, which means that the slope of p is greater than the slope of n:
Attachment:
2.PNG
So in both cases the slope of p is greater than the slope of n. Sufficient.
Answer: C.
So talking about the case with negative slopes here. OK so line 'n' is steeper hence it has a higher absolute value for slope right? Then because it is more negative then it is in fact smaller than the slope of line p.
So here we are saying that the slope is treated just as any number, which means considering its sign
Eg. Slope of an horizontal line will be higher than a negative slope right? Just cause if one does it algebraically one will encounter the comparison and absolute values are not used at all in slope formulae as far as I'm aware
Re: Lines n and p lie in the xy-plane. Is the slope of line n [#permalink]
02 Sep 2014, 02:54
Bunuel wrote:
BANON wrote:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p ?
(1) Lines n and p intersect at the point (5 , 1). (2) The y-intercept of line n is greater than the y-intercept of line p.
Algebraic approach:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p?
We have two lines: \(y_n=m_1x+b_1\) and \(y_p=m_2x+b_2\). Q: \(m_1<m_2\) true?
(1) Lines n and p intersect at the point (5,1) --> \(1=5m_1+b_1=5m_2+b_2\) --> \(5(m_1-m_2)=b_2-b_1\). Not sufficient.
(2) The y-intercept of line \(n\) is greater than the y-intercept of line \(p\) --> y-intercept is value of \(y\) for \(x=0\), so it's the value of \(b\) --> \(b_1>b_2\) or \(b_2-b_1<0\). Not sufficient.
(1)+(2) \(5(m_1-m_2)=b_2-b_1\), as from (2) \(b_2-b_1<0\) (RHS), then LHS (left hand side) also is less than zero \(5(m_1-m_2)<0\) --> \(m_1-m_2<0\) --> \(m_1<m_2\). Sufficient.
Re: Lines n and p lie in the xy-plane. Is the slope of line n [#permalink]
02 Sep 2014, 03:25
Expert's post
Sidhrt wrote:
Bunuel wrote:
BANON wrote:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p ?
(1) Lines n and p intersect at the point (5 , 1). (2) The y-intercept of line n is greater than the y-intercept of line p.
Algebraic approach:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p?
We have two lines: \(y_n=m_1x+b_1\) and \(y_p=m_2x+b_2\). Q: \(m_1<m_2\) true?
(1) Lines n and p intersect at the point (5,1) --> \(1=5m_1+b_1=5m_2+b_2\) --> \(5(m_1-m_2)=b_2-b_1\). Not sufficient.
(2) The y-intercept of line \(n\) is greater than the y-intercept of line \(p\) --> y-intercept is value of \(y\) for \(x=0\), so it's the value of \(b\) --> \(b_1>b_2\) or \(b_2-b_1<0\). Not sufficient.
(1)+(2) \(5(m_1-m_2)=b_2-b_1\), as from (2) \(b_2-b_1<0\) (RHS), then LHS (left hand side) also is less than zero \(5(m_1-m_2)<0\) --> \(m_1-m_2<0\) --> \(m_1<m_2\). Sufficient.
Re: Lines n and p lie in the xy-plane. Is the slope of line n [#permalink]
02 Sep 2014, 03:33
Sidhrt wrote:
Bunuel wrote:
BANON wrote:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p ?
(1) Lines n and p intersect at the point (5 , 1). (2) The y-intercept of line n is greater than the y-intercept of line p.
Algebraic approach:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p?
We have two lines: \(y_n=m_1x+b_1\) and \(y_p=m_2x+b_2\). Q: \(m_1<m_2\) true?
(1) Lines n and p intersect at the point (5,1) --> \(1=5m_1+b_1=5m_2+b_2\) --> \(5(m_1-m_2)=b_2-b_1\). Not sufficient.
(2) The y-intercept of line \(n\) is greater than the y-intercept of line \(p\) --> y-intercept is value of \(y\) for \(x=0\), so it's the value of \(b\) --> \(b_1>b_2\) or \(b_2-b_1<0\). Not sufficient.
(1)+(2) \(5(m_1-m_2)=b_2-b_1\), as from (2) \(b_2-b_1<0\) (RHS), then LHS (left hand side) also is less than zero \(5(m_1-m_2)<0\) --> \(m_1-m_2<0\) --> \(m_1<m_2\). Sufficient.
Re: Lines n and p lie in the xy-plane. Is the slope of line n [#permalink]
02 Sep 2014, 03:54
Expert's post
Sidhrt wrote:
Sidhrt wrote:
Bunuel wrote:
Algebraic approach:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p?
We have two lines: \(y_n=m_1x+b_1\) and \(y_p=m_2x+b_2\). Q: \(m_1<m_2\) true?
(1) Lines n and p intersect at the point (5,1) --> \(1=5m_1+b_1=5m_2+b_2\) --> \(5(m_1-m_2)=b_2-b_1\). Not sufficient.
(2) The y-intercept of line \(n\) is greater than the y-intercept of line \(p\) --> y-intercept is value of \(y\) for \(x=0\), so it's the value of \(b\) --> \(b_1>b_2\) or \(b_2-b_1<0\). Not sufficient.
(1)+(2) \(5(m_1-m_2)=b_2-b_1\), as from (2) \(b_2-b_1<0\) (RHS), then LHS (left hand side) also is less than zero \(5(m_1-m_2)<0\) --> \(m_1-m_2<0\) --> \(m_1<m_2\). Sufficient.
Re: Lines n and p lie in the xy-plane. Is the slope of line n [#permalink]
20 Sep 2015, 02:14
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Lines n and p lie in the xy-plane. Is the slope of line n [#permalink]
06 Jan 2016, 12:43
BANON wrote:
Lines n and p lie in the xy-plane. Is the slope of line n less than the slope of line p ?
(1) Lines n and p intersect at the point (5 , 1). (2) The y-intercept of line n is greater than the y-intercept of line p.
Hi math experts, I've used the following approach to derive on the correct answer. Would appreciate your input.
(1) It's not possible to calculate the slope with only one point given. Also it's an intersection point which satisfies the equations of both lines. Not Sufficient (2) You cannot determine a slope, with info about y-intercept, one can manipulate randomply the x-line intersection and get diff. results regarding slopes. (1)+(2) We have point (5,1) and info about y intercept. By y-intercept x=0 and we know that line n has a greater y-intercept: Case 1 +ve: Line n (0, 3) and Line p (0, 2) --> Slope n \(= \frac{1-3}{5}=-\frac{2}{5}\), Slope p\(=\frac{1-2}{5}=-\frac{1}{5}\), So Slope p > Slope n Case 2 -ve: Line n (0, -2) and Line p (0, -3) --> Slope n \(= \frac{1-(-2)}{5}=\frac{3}{5}\), Slope p=\(\frac{1-(-3)}{5}=\frac{4}{5}\) Again Slope p > Slope n
Answer C _________________
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