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

Question

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

CareerGeek · Answer

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

ajaygaur319 · Answer

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 .

Gauravji21 · 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.

bumpbot · Answer

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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

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