Hello everyone

This is potentially a tricky question, as some of you have found. I agree with mathaddict's answer: Q is $\displaystyle (-\tfrac12, -5)$, but I have an alternative solution that avoids nasty things like the distance formula. So here it is, if you care to look it over.

First notice that the points that are equidistant from B and C lie on the perpendicular bisector of BC. In the attached diagram, that's MQ where M $\displaystyle (0, -3)$ is the mid-point of BC.

Now the gradient of BC $\displaystyle = -\frac14 \Rightarrow$ gradient of MQ $\displaystyle = 4$. So the equation of MQ is$\displaystyle y = 4x - 3$

which agrees with mathaddict's result.

Next, let's look at the bisector of the angle BAC, the line AQ. It would be nice if you could add the gradients of BA and AC together and divide by 2 to get the gradient of AQ, but (as I hope ukorov has now realised) you can't! But you can do it fairly easily like this:

Let $\displaystyle \angle BAC = \theta$. Then, from the coordinates of A and B, and the fact that AC is vertical:$\displaystyle \tan \theta = \frac{8}{6}=\frac43$.

And $\displaystyle \angle QAC = \tfrac12\theta \Rightarrow$ gradient of QA $\displaystyle = \cot\tfrac12\theta = m$, say.

Now $\displaystyle \tan\theta =\frac43= \frac{2\tan\tfrac12\theta}{1-\tan^2\tfrac12\theta}$ $\displaystyle \Rightarrow\frac43= \frac{2\left(\dfrac1m\right)}{1-\left(\dfrac1m\right)^2}$$\displaystyle =\frac{2m}{m^2-1}$

$\displaystyle \Rightarrow 4m^2 -4 = 6m$

$\displaystyle \Rightarrow 2m^2-3m-2=0$

$\displaystyle \Rightarrow (2m+1)(m-2)=0$

$\displaystyle \Rightarrow m = 2$ (the other root representing the gradient of the line perpendicular to this)

So the equation of AQ is$\displaystyle y-4=2(x-4)$

i.e. $\displaystyle y = 2x-4$

This meets MQ where$\displaystyle 2x-4=4x-3$

i.e. at the point $\displaystyle (-\tfrac12, -5)$.

Grandad