Data Sufficiency

Line q is defined by the equation y = mx + b, where m < 0. Does line q pass through (5,4)?

(1) When it is reflected around the x-axis, line q passes through (3,-4)

(2) When it is reflected around the y-axis, line q passes through (-5,3)

[spoiler]OA: D Source: Grockit[/spoiler]

rahulvsd wrote:Line q is defined by the equation y = mx + b, where m < 0. Does line q pass through (5,4)?

(1) When it is reflected around the x-axis, line q passes through (3,-4)

(2) When it is reflected around the y-axis, line q passes through (-5,3)

[spoiler]OA: D Source: Grockit[/spoiler]

m < 0

Plugging in 5 for x and 4 for y: 4 = 5m + b

So, converting the question to a formula...

b = 4 - 5m?

Part 1
Reflecting across the x-axis means that any point (i,j) on the reflected line equates to (i,-j) on the original line. This is because there is no change in the horizontal aspect of the point, but the vertical placement of the point shares x = 0 as its midpoint with its reflection.

Therefore, the point (3,4) exists on the original line. Plugging in for x and y, we get the following:

4 = 3m + b
b = 4 - 3m

This clearly answers the question (NO), since 3m can't equal 5m if m < 0. SUFFICIENT


Part 2
As with the previous part, it is easy to find the original point. This time (i,j) becomes (-i,j).

Therefore, the point (5,3) exists on the original line.
3 = 5m + b
b = 3 - 5m

Again, we know that 3 is not equal to 4, so we are able to answer the question (NO). SUFFICIENT

D

rahulvsd wrote:Line q is defined by the equation y = mx + b, where m < 0. Does line q pass through (5,4)?

(1) When it is reflected around the x-axis, line q passes through (3,-4)

(2) When it is reflected around the y-axis, line q passes through (-5,3)

[spoiler]OA: D Source: Grockit[/spoiler]

Few things must be noted about the reflection around axes:

1. If the mirror line is the x-axis, just change each (x, y) into (x, -y).

2. If the mirror line is the y-axis, just change each (x, y) into (-x, y).

The line q will pass through (5, 4) only if 4 = 5 m + b, hence the rephrased question is:

Is b = 4 = 5 m?

(1) When it is reflected around the x-axis, line q passes through (3, -4). This means that point (3, 4) is on line y = m x + b, hence 4 = 3 m + b or b = 4 - 3 m. In case m ≠ 0, b = 4 - 3 m proves that b ≠ 4 = 5 m. Sufficient

(2) When it is reflected around the y-axis, line q passes through (-5, 3). This means that point (5, 3) is on line y = m x + b, hence 3 = 5 m + b or b = 3 - 5 m. Since 3 is not 4, it proves that b ≠ 4 = 5 m. [spoiler]Sufficient

Take D[/spoiler]