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If Line k in the xy-plane has equation y = mx + b, where m

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If Line k in the xy-plane has equation y = mx + b, where m  [#permalink]

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New post 02 Sep 2010, 12:52
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If Line k in the xy-plane has equation y = mx + b, where m and b are constants, what is the slope of k?

(1) k is parallel to the line with equation y = (1-m)x + b +1.
(2) k intersects the line with equation y = 2x + 3 at the point (2, 7).
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If Line k in the xy-plane has equation y = mx + b, where m  [#permalink]

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New post 26 Mar 2012, 14:15
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If Line k in the xy-plane has equation y = mx + b, where m  [#permalink]

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New post 26 Mar 2012, 13:56
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If Line k in the xy-plane has equation y = mx + b, where m and b are constants, what is the slope of k?

(1) k is parallel to the line with equation y = (1-m)x + b +1.
(2) k intersects the line with equation y = 2x + 3 at the point (2, 7)

I have a doubt in stmnt (2). K intersects the line with equation y=2(x)+3 at point (2,7) means for the line y=2(x)+3 slope is 2,but with this we cannot able to find slope of the line K. is my reasoning is right. please explain
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If Line k in the xy-plane has equation y = mx + b, where m  [#permalink]

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If Line k in the xy-plane has equation y = mx + b, where m and b are constants, what is the slope of k?

\(y=mx+b\) is called point-intercept form of equation of a line. Where: \(m\) is the slope of the line; \(b\) is the y-intercept of the line; \(x\) is the independent variable of the function \(y\).

So we are asked to find the value of \(m\).

(1) k is parallel to the line with equation y = (1-m)x + b +1 --> parallel lines have the same slope --> slope of this line is \(1-m\), so \(1-m=m\) --> \(m=\frac{1}{2}\). Sufficient.

(2) k intersects the line with equation y = 2x + 3 at the point (2, 7) --> so line k contains the point (2,7) --> \(7=2m+b\) --> can not solve for \(m\). Not sufficient.

Answer: A.
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Re: coordinate geometry  [#permalink]

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New post 26 Mar 2012, 14:14
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Quote:
If Line k in the xy-plane has equation y = mx + b, where m and b are constants, what is
the slope of k?
(1) k is parallel to the line with equation y = (1-m)x + b +1.
(2) k intersects the line with equation y = 2x + 3 at the point (2, 7)

I have a doubt in stmnt (2). K intersects the line with equation y=2(x)+3 at point (2,7) means for the line y=2(x)+3 slope is 2,but with this we cannot able to find slope of the line K. is my reasoning is right. please explain


My 2 cents:

Questions is asking you what is the value of M

so

1) Since k and this line with slope (1-m) are parallel

then m = 1 - m or m = 1/2

thus it is sufficient

2) we know that the point 2, 7 is on the line line y = mx + b

so 7 = 2m + b

But we know nothing else so we cannot solve for m thus it is insufficient
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Re: Coordinate Geometry 2  [#permalink]

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New post 07 Jun 2012, 00:54
For line K, y=mx+b implies slope=m

From Statement-1,

As slopes of parallel lines are equal, m = (1-m)

m=1/2
Statement-1 is sufficient.

From statement-2,

line K passes through (2,7)
7=2.m + b
m = (7-b)/2
Not sufficient.

Ans:A
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If Line k in the xy-plane has equation y = mx + b, where m  [#permalink]

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New post 29 Mar 2019, 13:51
Hi chetan2u, VeritasKarishma, Bunuel, amanvermagmat, gmatbusters,

I did not understood the bold part in below question, can any please assist.

Bunuel wrote:
If Line k in the xy-plane has equation y = mx + b, where m and b are constants, what is the slope of k?

\(y=mx+b\) is called point-intercept form of equation of a line. Where: \(m\) is the slope of the line; \(b\) is the y-intercept of the line; \(x\) is the independent variable of the function \(y\).

So we are asked to find the value of \(m\).

(1) k is parallel to the line with equation y = (1-m)x + b +1 --> parallel lines have the same slope --> slope of this line is \(1-m\), so \(1-m=m\) --> \(m=\frac{1}{2}\). Sufficient.

(2) k intersects the line with equation y = 2x + 3 at the point (2, 7) --> so line k contains the point (2,7) --> \(7=2m+b\) --> can not solve for \(m\). Not sufficient.

Answer: A.

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Re: If Line k in the xy-plane has equation y = mx + b, where m  [#permalink]

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New post 29 Mar 2019, 23:05
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Hi

I couldd not get your query, the principle is as below:
Image
Image

Attachment:
1.jpg
1.jpg [ 20.17 KiB | Viewed 2510 times ]

Attachment:
2.PNG
2.PNG [ 18.18 KiB | Viewed 2516 times ]


Gmatprep550 wrote:
Hi chetan2u, VeritasKarishma, Bunuel, amanvermagmat, gmatbusters,

I did not understood the bold part in below question, can any please assist.

Bunuel wrote:
If Line k in the xy-plane has equation y = mx + b, where m and b are constants, what is the slope of k?

\(y=mx+b\) is called point-intercept form of equation of a line. Where: \(m\) is the slope of the line; \(b\) is the y-intercept of the line; \(x\) is the independent variable of the function \(y\).

So we are asked to find the value of \(m\).

(1) k is parallel to the line with equation y = (1-m)x + b +1 --> parallel lines have the same slope --> slope of this line is \(1-m\), so \(1-m=m\) --> \(m=\frac{1}{2}\). Sufficient.

(2) k intersects the line with equation y = 2x + 3 at the point (2, 7) --> so line k contains the point (2,7) --> \(7=2m+b\) --> can not solve for \(m\). Not sufficient.

Answer: A.

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Re: If Line k in the xy-plane has equation y = mx + b, where m  [#permalink]

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New post 30 Mar 2019, 00:19
gmatbusters wrote:
Hi

I couldd not get your query, the principle is as below:
Image
Image

Attachment:
1.jpg

Attachment:
2.PNG


Gmatprep550 wrote:
Hi chetan2u, VeritasKarishma, Bunuel, amanvermagmat, gmatbusters,

I did not understood the bold part in below question, can any please assist.

Bunuel wrote:
If Line k in the xy-plane has equation y = mx + b, where m and b are constants, what is the slope of k?

\(y=mx+b\) is called point-intercept form of equation of a line. Where: \(m\) is the slope of the line; \(b\) is the y-intercept of the line; \(x\) is the independent variable of the function \(y\).

So we are asked to find the value of \(m\).

(1) k is parallel to the line with equation y = (1-m)x + b +1 --> parallel lines have the same slope --> slope of this line is \(1-m\), so \(1-m=m\) --> \(m=\frac{1}{2}\). Sufficient.

(2) k intersects the line with equation y = 2x + 3 at the point (2, 7) --> so line k contains the point (2,7) --> \(7=2m+b\) --> can not solve for \(m\). Not sufficient.

Answer: A.


Hi gmatbusters

Apologies for not highlighting proper issue,

I was not able to get following part.


slope of this line is 1−m1−m, so 1−m=m1−m=m


Even if slope is m then also how 1-m = m? I feel that "m" mentioned in equation is different.
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Re: If Line k in the xy-plane has equation y = mx + b, where m  [#permalink]

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New post 29 Jul 2019, 22:11
Parallel lines have same slopes buddy

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Re: If Line k in the xy-plane has equation y = mx + b, where m   [#permalink] 29 Jul 2019, 22:11
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