GMATPrepNow wrote:
Line J: 4x - 7 = 6y
Line K: 6y + 3x = -2
Line L: 2x = 5 - y
In the x-y coordinate plane, lines J, K and L are defined by the above equations.
Point B is the point of intersection of lines J and K. Point C is the point of intersection of lines K and L.
What is the slope of line segment BC?
A) -3/2
B) -2/3
C) -1/2
D) 1/2
E) 2/3
Draw the lines approximately in the x-y coordinate with the help of the slopes. Refer to the figure in attachment below.
J : y=\(\frac{2}{3}\)x-\(\frac{7}{6}\)
K : y=\(\frac{-x}{2}\)-\(\frac{1}{3}\)
L : y=-2x+5
Point B is the point of intersection of lines J and K. Point C is the point of intersection of lines K and L.
Hence BC lies on line K, which has a slope of \(\frac{-1}{2}\).
Answer \(\frac{-1}{2}\) (C).
Attachments
Untitled.jpg [ 12.67 KiB | Viewed 1239 times ]
_________________
Please give kudos, if you like my post
When the going gets tough, the tough gets going...