In the rectangular coordinate system, line k passes through the point (n, −1). Is the slope of line k greater than zero?
(1) Line k passes through the origin.
(2) Line k passes through the point (1, n + 2).
This question clearly tests the understanding of the slope of the line.
If a line passes through the points (x1,y1) and (x2,y2) , then the slope of the line is \(\frac{(y2-y1)}{(x2-x1)}\).
Also, a few traps are hidden in the question. Here it's given that line k passes through the point (n,-1). Remember,
'n' could be any real number i.e positive, negative, or zero. We cannot assume that n is a positive value.
We are asked to find
if the slope of line K is positive or not.
(1) Line k passes through the origin.
It's already given in the question that the line passes through (n,-1). From St1, it is given the line passes through origin i.e (0,0).
Since the line passes through (0,0)and (n,-1), the slope of line k =\( \frac{(y2-y1)}{(x2-x1)}\) = \(\frac{(-1-0)}{(n-0)}\) = \(\frac{-1}{n}\)
Is the slope of line k is positive? It depends on the value of n.
If n is a negative value, the slope of the line is positive and if n is a positive value, the slope will be negative. Since we don't have a definite YES/ NO to the Question stem, Statement 1 alone is not sufficient.
(2) Line k passes through the point (1, n + 2).
We know about the 2 points the line k passes through i.e (n,-1) and (1, n + 2)
Slope of the line k = \(\frac{(n+2 --1)}{(1-n)}\)= \(\frac{(n+3)}{(1-n)}\).
Here 'n' will determine the sign of the slope in this case.
For example: n= -1, Slope = -1+3/1--1 = 2/2 = 1 ==> Slope is positive
n=2, slope = 2+3/1-2 = 5/-1 = -5 ==> Slope is negative
Based on the value of n, the slope of the line k could be positive or negative. Hence Statement 2 alone is not sufficient.
The next step is to combine both statements.
We figured out that the slope of line k from St 1 is \(\frac{ -1}{n} \). From statement 2, we found that the slope of the same line is \( \frac{(n+3)}{(1-n)}\)
Since both the slopes represent the same line k, we can equate both.
\(\frac{ -1}{n} \) = \( \frac{(n+3)}{(1-n)}\)
n-1 = \(n^2\) + 3n
\(n^2\) + 2n + 1 = 0
\((n+1)^2\) = 0
n+1 =0 ===> n =-1
On solving, we found the value of n as -1.
Therefore , the slope of line k = \(\frac{ -1}{n} \) = -1/-1 = 1 (positive)
Is the slope of line k greater than zero? YESWe are getting a definite answer to the Question stem by combining both statements. Hence,
Option C is the correct answer.Thanks,
Clifin J Francis,
GMAT Mentor
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