enigma123 wrote:

The line represented by the equation y = 4 – 2x is the perpendicular bisector of line segment RP. If R has the coordinates (4, 1), what are the coordinates of point P?

(A) (–4, 1)

(B) (–2, 2)

(C) (0, 1)

(D) (0, –1)

(E) (2, 0)

If the line y = 4 - 2x is the perpendicular bisector of line segment RP, then it’s perpendicular to RP at the midpoint of RP. Recall that two lines are perpendicular if their slopes are negative reciprocals of each other. Since the line has a slope of -2, the line segment should have a slope of ½. Thus, let’s first determine which answer choice can be point P so that the slope of RP is ½. We’ll use the slope formula: m = (y2 - y1)/(x2 - x1):

A) (-4, 1)

m = (1 - 1)/(4 - (-4)) = 0

Point P can’t be (-4, 1).

B) (-2, 2)

m = (1 - 2)/(4 - (-2)) = -1/6

Point P can’t be (-2, 2).

C) (0, 1)

m = (1 - 1)/(4 - 0) = 0

Point P can’t be (0, 1).

D) (0, -1)

m = (1 - (-1))/(4 - 0) = 2/4 = 1/2

Point P could be (0, -1).

E) (2, 0)

m = (1 - 0)/(4 - 2) = 1/2

Point P could be (2, 0).

We see that point P could be either (0, -1) or (2, 0), since either one will make RP’s slope ½. Next let’s determine the midpoint of RP if P is either (0, -1) or (2, 0). We use the midpoint formula: ((x1 + x2)/2 , (y1 + y2)/2)). Recall that R = (4,1).

If P = (0, -1), then the midpoint of RP = ((4 + 0)/2, (1 + (-1))/2) = (2, 0).

If P = (2, 0), then the midpoint of RP = ((4 + 2)/2, (1 + 0)/2) = (3, 1/2).

Recall that the line y = 4 - 2x has to include the midpoint of RP. In other words, the midpoint of RP is a point on the line, and hence its coordinates will satisfy the equation of the line.

If P = (0, -1) and the midpoint of RP is (2, 0), is 0 = 4 - 2(2)? The answer is yes, since 0 = 0. We can see that choice D is the correct choice. However, let’s also show that choice E is not the correct choice:

If P = (2, 0) and the midpoint of RP is (3, 1/2), is 1/2 = 4 - 2(3)? The answer is no, since 1/2 ≠ -2.

Alternate Solution:

Since the line segment RP is perpendicular to y = 4 - 2x, the line containing RP must have a slope of 1/2, since the slopes of perpendicular lines are negative reciprocals of each other. Then, the line containing the line segment RP must be of the form y = (1/2)x + b, for some real number b. Since this line contains the point R, its equation must be satisfied when we substitute x = 4 and y = 1; therefore 1 = (1/2)(4) + b. Then, 1 = 2 + b, and thus b = -1.

We know the line that contains the line segment RP has an equation y = (1/2)x - 1. Let’s find the common point of this line with y = 4 - 2x. We set (1/2)x - 1 = 4 - 2x and solve for x:

(1/2)x -1 = 4 - 2x

(5/2)x = 5

x = 2

So, the x-coordinate of the common point is 2. We can substitute x = 2 in either equation to find the y-coordinate: y = 4 - 2(2) = 0. So, the common point is (2, 0).

Note that this common point is the midpoint of R and P. Therefore, if we let P = (a, b), we must have:

(a + 4)/2 = 2 and (b + 1)/2 = 0. We find that a = 0 and b = -1. Thus, P = (0, -1).

Answer: D

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