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Difficulty: HardDistance and Midpoint Formulas

In the standard (x,y)(x, y) coordinate plane, the midpoint of the line segment with endpoints A(2,a)A(2, a) and B(b,−3)B(b, -3) lies on the line y=3x−4y = 3x - 4. If the distance between AA and BB is 525\sqrt{2} and a>0a > 0, what is the value of a+ba + b?

  1. A
    13
  2. B
    -11
  3. 5Answer
  4. D
    3
  5. E
    1

Answer

The value of a+ba + b is 55.
The midpoint of the segment with endpoints (2,a)(2, a) and (b,−3)(b, -3) is (2+b2,a−32)\left(\frac{2+b}{2}, \frac{a-3}{2}\right). Substituting these coordinates into the line y=3x−4y = 3x - 4 gives a−32=3(2+b2)−4\frac{a-3}{2} = 3\left(\frac{2+b}{2}\right) - 4, which simplifies to a=3b+1a = 3b + 1. The squared distance between the endpoints is (b−2)2+(−3−a)2=(52)2=50(b-2)^2 + (-3-a)^2 = (5\sqrt{2})^2 = 50. Substituting a=3b+1a = 3b + 1 into the distance equation yields (b−2)2+(3b+4)2=50(b-2)^2 + (3b+4)^2 = 50, which simplifies to the quadratic equation b2+2b−3=0b^2 + 2b - 3 = 0. Factoring gives (b+3)(b−1)=0(b+3)(b-1) = 0, so b=1b = 1 or b=−3b = -3. The constraint a>0a > 0 means a=3(1)+1=4a = 3(1) + 1 = 4 is the only valid solution. Therefore, the sum is 4+1=54 + 1 = 5.

Step-by-Step Solution

1
Find the midpoint of the segment ABAB using the midpoint formula.
M=(2+b2,a−32)M = \left(\frac{2+b}{2}, \frac{a-3}{2}\right)
The midpoint is defined as the average of the x-coordinates and the average of the y-coordinates.
2
Substitute the midpoint coordinates into the equation of the line y=3x−4y = 3x - 4 to relate aa and bb.
a=3b+1a = 3b + 1
Since the midpoint lies on the line, its coordinates must satisfy the line's equation.
3
Use the distance formula to set up an equation for the distance between AA and BB.
(b−2)2+(a+3)2=50(b-2)^2 + (a+3)^2 = 50
The squared distance between (2,a)(2, a) and (b,−3)(b, -3) is (52)2=50(5\sqrt{2})^2 = 50.
4
Substitute the relation a=3b+1a = 3b + 1 into the distance equation and solve for bb.
b=1b = 1 or b=−3b = -3
This yields the quadratic equation b2+2b−3=0b^2 + 2b - 3 = 0, which factors as (b+3)(b−1)=0(b+3)(b-1) = 0.
5
Apply the constraint a>0a > 0 to find the correct values of aa and bb.
a=4a = 4 and b=1b = 1
If b=−3b = -3, then a=3(−3)+1=−8a = 3(-3) + 1 = -8, which is not greater than 00. If b=1b = 1, then a=3(1)+1=4>0a = 3(1) + 1 = 4 > 0.
6
Calculate the sum a+ba + b.
4+1=54 + 1 = 5
To find the final requested value.

Key Concept

Using the distance and midpoint formulas in combination with linear equations to solve for unknown coordinates.
Estimated Time:2m 30s
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Topics

Distance and Midpoint Formulas

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