Question

Difficulty: HardDistance and Midpoint Formulas

In the standard (x,y)(x, y) coordinate plane, a line segment has midpoint M(4,1)M(4, 1) and one endpoint P(x,5)P(x, 5). If the total length of the line segment is 1010 units, and x<4x < 4, what is the value of xx?

Answer: 1

Answer

1
The midpoint MM divides the line segment into two equal segments, meaning the distance from endpoint PP to midpoint MM is half of the total length of the segment, 102=5\frac{10}{2} = 5 units. Applying the distance formula between P(x,5)P(x, 5) and M(4,1)M(4, 1) gives (x4)2+(51)2=5\sqrt{(x - 4)^2 + (5 - 1)^2} = 5. Squaring both sides yields (x4)2+16=25(x - 4)^2 + 16 = 25, which simplifies to (x4)2=9(x - 4)^2 = 9. Solving for xx gives x4=3x - 4 = 3 or x4=3x - 4 = -3, meaning x=7x = 7 or x=1x = 1. The problem specifies that x<4x < 4, so the only valid value is 11.

Step-by-Step Solution

1
Find the distance from the endpoint to the midpoint.
The distance PMPM is 55 units.
The midpoint divides the line segment into two equal parts, so the distance from any endpoint to the midpoint is half of the total length of the segment: 102=5\frac{10}{2} = 5.
2
Set up the distance formula equation for the segment PMPM.
(x4)2+(51)2=5\sqrt{(x - 4)^2 + (5 - 1)^2} = 5
Using the coordinates of P(x,5)P(x, 5) and M(4,1)M(4, 1) with the distance formula d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.
3
Simplify the equation and square both sides.
(x4)2+16=25(x - 4)^2 + 16 = 25
Calculating (51)2=16(5 - 1)^2 = 16 and squaring both sides of the equation to eliminate the radical.
4
Isolate the squared term and solve for xx.
x=7x = 7 or x=1x = 1
Subtracting 1616 from both sides gives (x4)2=9(x - 4)^2 = 9. Taking the square root gives x4=3x - 4 = 3 or x4=3x - 4 = -3, yielding x=7x = 7 or x=1x = 1.
5
Apply the constraint x<4x < 4.
x=1x = 1
Since the question states that x<4x < 4, we choose x=1x = 1 instead of x=7x = 7.

Key Concept

The midpoint divides a segment into two segments of equal length, and the distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by (x2x1)2+(y2y1)2\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}.
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