The perpendicular bisected of the line segment joining $P(1,4)$ and $Q(k,3)$ has $y$-intercept $4$. Then the possible value of k is

Question

Slope of line $PQ$ is $$m=\frac{1}{1-k}$$ The slope perpendicular to it will be $k-1$

Since the line is a bisector of PQ it will pass through $(\frac{1+k}{2},\frac 72)$ Then $$y=mx+c$$ $$\frac 72 =(k-1)(\frac {k+1}{2})+4$$ $$\frac{-1}{2}=\frac{k^2-1}{2}$$ $$k=0$$ But the answer given is $-4$. What’s going wrong?

Answer 1

Nothing. Your answer is fine. Note that, when $k=0$, $Q=(0,3)$. Since $P=(1,4)$, the slope of the line passing through $P$ and $Q$ is $1$, and therefore the slope of the perpendicular bisector of $P$ and $Q$ is $-1$. But the line passing through $\left(\frac12,\frac72\right)$ with slope $-1$ clearly passes through $(0,4)$ too.

Of course, this doesn't prove that the answer $k=-4$ is wrong (although it is wrong), but it shows that asserting that that's the only answer cannot be right.

Answer 2

$$(\frac{k+1}{2},\frac{7}{2})+t(1,k-1)=(0,4)$$ $$\begin{cases} k+1+2t=0,\\ 7-t(k-1)=4 \end{cases}$$ Since this is a system of linear equations, it has $0,\,1\hbox{ or }\infty$ solutions. $$k=0,\,t=-\frac12$$

Answer 3

Option:

The perpendicalur bisector intersects the $y$-axis in $(0,4)$.

A circle with center $(0,4)$ passes through $P(1,4)$ and $Q(k,3)$.

Radius of this circle: $r=\sqrt{1^2+0^2}=1$;

Hence

$1^2=k^2+(4-3)^2= k^2+1^2$;

$k=0$;