In the xy-plane, if line k has negative slope and passes

Drawing method:

(1) Line k passes through the point (0, -2) and has negative slope. Line k can be something like this:
So you can see that x intercept must be negative. Sufficient.

(2) The y-intercept of line k is negative and line has negative slope. Line k can be something like this:
Again you can see that when y-intercept is negative and slope is negative, then x-intercept must be negative too. Sufficient.

Answer: D.

Algebraic approach:

Equation of a line in point intercept form is \(y=mx+b\), where: \(m\) is the slope of the line; \(b\) is the y-intercept of the line (the value of \(y\) for \(x=0\)); \(x\) is the independent variable of the function \(y\).

X-intercept is the value of \(x\) for \(y=0\) --> \(x_{intercept}=-\frac{b}{m}\).

Question: is \(-\frac{b}{m}>0\). As given that slope is negative (\(m<0\)), basically the question becomes is \(b>0\)?

(1) Line k passes through the point (0, -2) --> \(-2=0*m+b\) --> \(b=-2<0\). Answer to the question is NO. Sufficient.

(2) The y-intercept of line k is negative --> as y-intercept is value of \(b\), so we are directly told that \(b<0\). Answer to the question is NO. Sufficient.

Answer: D.

For more on this issue refer to the Coordinate Geometry chapter of the Math book (link in my signature).

Hope it helps.