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Hi chetan2u, VeritasKarishma, Bunuel, amanvermagmat, gmatbusters,

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

Bunuel
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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Hi

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



Attachment:
1.jpg
1.jpg [ 20.17 KiB | Viewed 34932 times ]
Attachment:
2.PNG
2.PNG [ 18.18 KiB | Viewed 34952 times ]

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

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

Bunuel
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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Bunuel
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.

Thanks! But is this implying that no matter the format of the equation, the term/number multiplying the x-value is the gradient? I was thrown off by '+ 1' in the equation.
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I have the same question! KarishmaB Bunuel could you please clarify?
adkor95
Bunuel
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.
Thanks! But is this implying that no matter the format of the equation, the term/number multiplying the x-value is the gradient? I was thrown off by '+ 1' in the equation.
­
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mollyweasley
I have the same question! KarishmaB Bunuel could you please clarify?
adkor95
Bunuel
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.
Thanks! But is this implying that no matter the format of the equation, the term/number multiplying the x-value is the gradient? I was thrown off by '+ 1' in the equation.
­
­For Coordinate Geometry check:

24. Coordinate Geometry



For more check:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.­
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adkor95
Bunuel
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.
Thanks! But is this implying that no matter the format of the equation, the term/number multiplying the x-value is the gradient? I was thrown off by '+ 1' in the equation.
­Yes, m, the co-efficient of x in the equation of this format y = mx + c  gives the slope of the line. 

y = 3x + 2
y = 3x - 1
Both have the same slope 3. They are parallel lines.

Check the concepts of Graphing here: 
https://youtu.be/3kX5UtvHGFg­
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