AIEEE 2008 (Offline)Question 1
The perpendicular bisector of the line segment joining $P(1,4)$ and $Q(k,3)$ has $y$-intercept $-4$. Then a possible value of $k$ is :
Points and Straight Lines
Solution
Midpoint $\left(\frac{1+k}{2},\frac72\right)$, slope of $PQ$ is $\frac{-1}{k-1}$, so the bisector is $y-\frac72=(k-1)\left(x-\frac{1+k}{2}\right)$. At $x=0$: $\frac72-\frac{k^2-1}{2}=-4\Rightarrow k^2=16$, so $k=-4$ is possible.