Question

The coordinates of P and Q are (0, 4) and (a, 6), respectively, where a is a non-zero integer. R is the midpoint of PQ. The perpendicular bisector of PQ cuts the X-axis at the point S(b, 0). For how many integer values of a is b also an integer?

• A. 4
• B. 3
• C. 2
• D. 1

Hint:

Write b in terms of a using the perpendicular bisector equation, then check which integer values of a keep b a whole number.

Answer 1

The Correct Option is A

Step 1: Find the midpoint R and the slope of PQ.
P is (0, 4) and Q is (a, 6). The midpoint of two points is found by averaging their coordinates, so \( R = \left(\frac{0+a}{2}, \frac{4+6}{2}\right) = \left(\frac{a}{2}, 5\right) \).
The slope of PQ is \( \frac{6-4}{a-0} = \frac{2}{a} \).

Step 2: Write the equation of the perpendicular bisector.
A line perpendicular to PQ has a slope that is the negative reciprocal of PQ's slope, so its slope is \( -\frac{a}{2} \).
This perpendicular line passes through R, so its equation is \( y - 5 = -\frac{a}{2}\left(x - \frac{a}{2}\right) \).

Step 3: Find where this line crosses the X-axis.
On the X-axis, y = 0, so this point is exactly S(b, 0). Substituting y = 0 and x = b:
\[ -5 = -\frac{a}{2}\left(b - \frac{a}{2}\right) \]
Multiplying both sides by \(-1\) and simplifying, \( 5 = \frac{a}{2}b - \frac{a^2}{4} \), which rearranges to \( b = \frac{a}{2} + \frac{10}{a} \).

Step 4: Decide which integer values of a make b an integer.
Write \( b = \frac{a^2 + 20}{2a} \). For b to be a whole number, a must divide evenly into \(a^2 + 20\); since a always divides \(a^2\), this forces a to divide 20 as well. The integer divisors of 20 are \( \pm1, \pm2, \pm4, \pm5, \pm10, \pm20 \). Checking each one in \( b = \frac{a}{2} + \frac{10}{a} \): a = 2 gives b = 1 + 5 = 6 (integer); a = 10 gives b = 5 + 1 = 6 (integer); a = -2 gives b = -1 - 5 = -6 (integer); a = -10 gives b = -5 - 1 = -6 (integer). The remaining divisors (1, 4, 5, 20 and their negatives) each leave a leftover half, for example a = 1 gives b = 0.5 + 10 = 10.5, which is not an integer.

Why the other options are wrong:
3, 2 and 1 come from stopping the check early, for instance only counting the positive values of a and missing that a = -2 and a = -10 work just as well, or from wrongly excluding one of the four valid values.

Final Answer:
There are 4 integer values of a, namely -10, -2, 2 and 10, for which b is also an integer.
\[ \boxed{4} \]