Good question. (1) is sufficient because if a line has a negative slope (meaning it goes up and to the left), it will eventually reach Quadrant II (top left quadrant) no matter how far down & right you begin to draw the line. Remember that lines have an infinite length.

(2) gives us one point of the line which is not in Quadrant II. The line itself could be drawn to touch quadrant II, or to not enter it at all. So (2) is not sufficient.

A is the answer.


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-Patrick

Let the angle between K and x-axis be A.Then slope is tan(A)
From 1
tan(A)=-1/6 so 90> A>180..hence the line will pass through quadrant II...sufficient

From 2
y-intercept of K is -6 ..but this is not sufficient to comment whether slope is positive or negative..not sufficient

Ans option A

p.s its easy to understand this if we draw the lines.

Hi liferocks
Sorry,

tan(A)=-1/6 so 90> A>180..hence the line will pass through quadrant ....

This part is not clear to me.can you please elaborate.

Thaks & Regards

liferocks wrote:Let the angle between K and x-axis be A.Then slope is tan(A)
From 1
tan(A)=-1/6 so 90> A>180..hence the line will pass through quadrant II...sufficient

From 2
y-intercept of K is -6 ..but this is not sufficient to comment whether slope is positive or negative..not sufficient

Ans option A

p.s its easy to understand this if we draw the lines.

Adding on to that:

If the slope is negative the line has to pass through second & fourth quadrant!!