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When line k, expressed by the equation y = mx + b, is graphed on a coo

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When line k, expressed by the equation y = mx + b, is graphed on a coo [#permalink] New post   07 Feb 2020, 06:14
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When line k, expressed by the equation y = mx + b, is graphed on a coordinate axis, does k intersect the origin?

(1) m = 0

(2) b > 0
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Re: When line k, expressed by the equation y = mx + b, is graphed on a coo [#permalink] New post   09 Feb 2020, 02:26
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Bunuel wrote:
When line k, expressed by the equation y = mx + b, is graphed on a coordinate axis, does k intersect the origin?

(1) m = 0

(2) b > 0


A line will intersect the origin ONLY when its y-intercept is zero

(1) m = 0
--> The line k is parallel to x-axis and we cannot conclude anything about the y-intercept --> Insufficient

(2) b > 0
--> Since the y-intercept is positive, we can conclude that line k does not pass through the origin
--> A Definite NO --> Sufficient

Option B
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When line k, expressed by the equation y = mx + b, is graphed on a coo [#permalink] New post   09 Feb 2020, 06:34
Bunuel wrote:
When line k, expressed by the equation y = mx + b, is graphed on a coordinate axis, does k intersect the origin?

(1) m = 0

(2) b > 0


if the line y = mx + b passes through origin then, putting (0, 0) in the equation:
b = 0

1 --> m = 0, in this case y = b which cannot tell if the line is passing through origin, hence not sufficient
2 --> b > 0, This suggests that the y-intercept of the line is > 0 but this cannot tell if the line is passing through origin. e.g. y = 2x + 1 doesn't passed though origin. Hence not sufficient

Combining both statements, y = b and b > 0, hence the line is parallel to x-axis with b > 0 and the line cannot pass through origin. Hence, C is the correct answer.
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Re: When line k, expressed by the equation y = mx + b, is graphed on a coo [#permalink] New post   08 Sep 2020, 19:07
ajaygaur319 wrote:
Bunuel wrote:
When line k, expressed by the equation y = mx + b, is graphed on a coordinate axis, does k intersect the origin?

(1) m = 0

(2) b > 0


if the line y = mx + b passes through origin then, putting (0, 0) in the equation:
b = 0

1 --> m = 0, in this case y = b which cannot tell if the line is passing through origin, hence not sufficient
2 --> b > 0, This suggests that the y-intercept of the line is > 0 but this cannot tell if the line is passing through origin. e.g. y = 2x + 1 doesn't passed though origin. Hence not sufficient

Combining both statements, y = b and b > 0, hence the line is parallel to x-axis with b > 0 and the line cannot pass through origin. Hence, C is the correct answer.



Even if a line is not parallel to Y axis, it can avoid passing through the origin. example: an inclined line passing through a point lets say (10,20) in I quadrant and Y intercept of (0,5). This line will have y intercept other than zero. So, this line will not pass from origin.

Also, if m=0, y = b i.e a line parallel to X axis or X axis itself. In case, line is parallel to X axis, it will not intersect Origin but if it is X axis itself, it will intersect origin. hence Statement 1 is not sufficient.

Statement 2 is sufficient as b>0 which clearly indicates that y has got an intercept which i more than zero. So, it will not pass through origin.
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Re: When line k, expressed by the equation y = mx + b, is graphed on a coo [#permalink] New post   17 Feb 2023, 07:40
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Re: When line k, expressed by the equation y = mx + b, is graphed on a coo [#permalink]
17 Feb 2023, 07:40
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