In x-y co-ordinate plane, a line l passes through a point (4, p).

The stem gives us point (4,p) and asks the slope. We know that (4,p) will be somewhere on the vertical line x=4.

(1) Gives us the point (0,-1). We could pick any one of infinite points along the line x=4 to satisfy (4,p) and each of them would result in a different slope. So, does statement (1) alone give us enough information to determine the slope? No. BCE.

(2) Gives us the point (p+5,-2). We know that (p+5,-2) will be somewhere on the horizontal line y=-2. We could pick any point on x=4 to satisfy the stem and any point on y=-2 to satisfy statement (2). There are infinite combinations and infinite possible slopes. So, does statement (2) alone give us enough information to determine the slope? No. CE.

(1&2) We now need to find a line that contains all of (0,-1), (4,p), and (p+5,-2). We know that (4,p) resides on x=4 and (p+5,-2) resides on y=-2 and we also know that (0,-1) doesn't reside on either of those lines, so the only way for all three points to lie on the same line is for (4,p) and (p+5,-2) to be the point where x=4 and y=-2 intersect, which is at (4,-2). We now know that line l passes through both (0,-1) and (4,-2). Do both statements together give us enough information to answer the question? Yes. C.

Answer choice C.

For a slope of a line, I need any two points on the line.
\(Slope = \frac{(y2 - y1)}{(x2 - x1)}\)

I am given that the line passes through (4, p) but I don't know what p is.

(1) The y-intercept of line l is equal to -1.
I am given a point through which the line passes (0, -1). But I don't have any other point. I don't know the value of p. Not sufficient alone.

(2) Point (p + 5, -2) lies on the line l.

I have two points (4, p) and (p + 5, -2). So slope will be

\(\frac{-2 - p}{p + 5 - 4} = \frac{-p - 2}{p + 1}\)

Since I don't know the value of p, I can't get the slope. The reason I would check what slope is here is that both points have p. What if the p's were to get cancelled to give a fixed slope.

Not sufficient alone.

Using both together, now I have 3 points, 2 of which are in p. We can find slope in two ways and equate them.
Slope is
\( \frac{(p+1)}{(4-0)} = \frac{(-2 + 1)}{(p + 5)} \)
(I know that I will get a quadratic but I need to solve quadratics in DS because I may get same root)
\((p + 3)^2 = 0\)
\(p = -3\)

So line passes through (4, -3), (0, -1) and (2, -2).
Now, we can get the slope so both together are sufficient.


Answer (C)