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

Instead of writing the perpendicular bisector as a slope-intercept equation, let's build it using a direction vector, a slightly different route to the same relationship between a and b.

1. Direction of PQ: going from P(0, 4) to Q(a, 6), the direction vector is $(a, 2)$.
2. Perpendicular direction: a vector perpendicular to $(a, 2)$ is $(2, -a)$, since their dot product is $a(22) + 2(-a) = 0$.
3. Parametric line through the midpoint: the midpoint R is $\left(\frac{a}{2}, 5\right)$. Moving along the perpendicular direction by a parameter $t$, points on the bisector are $\left(\frac{a}{2} + 2t,\ 5 - at\right)$.
4. Find where y = 0: setting $5 - at = 0$ gives $t = \frac{5}{a}$.
5. Substitute back for x = b: $b = \frac{a}{2} + 2 \times \frac{5}{a} = \frac{a}{2} + \frac{10}{a}$, the same expression as before, confirming the two methods agree.

Now check which integers a make this a whole number. Since $\frac{10}{a}$ only contributes a whole-number style fraction alongside $\frac{a}{2}$ when a shares factors with 20, a must be a divisor of 20: the candidates are $\pm1, \pm2, \pm4, \pm5, \pm10, \pm20$. Testing each shows only $a = 2, 10, -2, -10$ give a whole-number b (each giving $b = 6$ or $b = -6$); every other divisor leaves a leftover half unit, for example $a = 4$ gives $b = 2 + 2.5 = 4.5$.

Let's summarize:

• The parametric (vector) method and the slope-intercept method both give $b = \frac{a}{2} + \frac{10}{a}$.
• Only 4 integer values of a, namely $-10, -2, 2, 10$, keep b a whole number.

So there are 4 integer values of a for which b is also an integer.