# Math Help - coordinate geometry

1. ## coordinate geometry

The vertices of a triangle are A(4,4) , B(-4,-2) and C(4,-4) respectively . Fidn the coordinate of the point which is equidistant from B and C and which lies on the angle bisector of angle BAC .

which lies on the angle bisector of angle BAC --- How do i use this part of the information ?

2. slope of AB = $\frac{6}{8} = \frac{3}{4}$
slope of AC = $\frac{4}{4} = 1$

let the bisector of angle BAC intersects with BC at D.
hence slope of AD = $(\frac{3}{4} + 1)(\frac{1}{2}) = \frac{7}{8}$

$\frac{y - 4}{x - 4} = \frac{7}{8}$
you get 8y = 7x + 4
multiply by 3 you get 24y = 21x + 12 ...(1)

Let E be the point lying on AD that is equidistant to both B and C.
Since BE = CE,
$\sqrt{(y + 2)^2 + (x + 4)^2} = \sqrt{(y + 4)^2 + (x + 4)^2}$
$(y + 2)^2 + (x + 4)^2 = (y + 4)^2 + (x + 4)^2$
$(y + 2)^2 = (y + 4)^2$
$y^2 + 4y + 4 = y^2 + 8y + 16$
In the end you get y = -3 ...(2)

Substitute (2) into (1) you get x = -4
so the point in question is (-4, -3)

3. Originally Posted by ukorov
slope of AB = $\frac{6}{8} = \frac{3}{4}$
slope of AC = $\frac{4}{4} = 1$

let the bisector of angle BAC intersects with BC at D.
hence slope of AD = $(\frac{3}{4} + 1)(\frac{1}{2}) = \frac{7}{8}$

$\frac{y - 4}{x - 4} = \frac{7}{8}$
you get 8y = 7x + 4
multiply by 3 you get 24y = 21x + 12 ...(1)

Let E be the point lying on AD that is equidistant to both B and C.
Since BE = CE,
$\sqrt{(y + 2)^2 + (x + 4)^2} = \sqrt{(y + 4)^2 + (x + 4)^2}$
$(y + 2)^2 + (x + 4)^2 = (y + 4)^2 + (x + 4)^2$
$(y + 2)^2 = (y + 4)^2$
$y^2 + 4y + 4 = y^2 + 8y + 16$
In the end you get y = -3 ...(2)

Substitute (2) into (1) you get x = -4
so the point in question is (-4, -3)
are you sure the gradient of AC is 1 ?

are you sure the gradient of AC is 1 ?
Oh....I read it wrong....C is (4, -4) and slope of AC is undefined
let me think...

5. I tried to work out the slope of AD by producing a right angled triangle ABX, with X lying on AC such that angle AXB is the right angle. The triangle is rotated $90^o$ clockwise as shown in attached figure.
slope of AB becomes - $\frac{4}{3}$
slope of AX (AC) becomes 0
now the slope of AD can be calculated as $-\frac{2}{3}$ (correct?)
therefore the slope of AD before rotation should be $\frac{3}{2}$then,
equation of AD should be 2y = 3x - 4 or 24y = 36x - 48

the calculation for value of y is not affected, ie. y = -3
and that should yield $x = -\frac{2}{3}$
hope there is no mistake this time

6. Hi ukorov , this is what i got .

Let Q=(x,y)

Given that BQ=QC

$\sqrt{(x+4)^2+(y+2)^2}=\sqrt{(x-4)^2+(y+4)^2}$

eventually simplify to $y=4x-3$ ---- 1

Draw 2 perpendiculars QR and QS which are equal because

$\triangle ARQ \cong \triangle ASQ$ (AAS)

Then use the perpendicular distance formula

$|\frac{x-4}{1}|=|\frac{-3x+4y-4}{\sqrt{3^2+4^2}}|$

$|5x-20|=|-3x+4y-4|$

$y=2x-4$ ---- 2

Solve the simultaneous equation , i got Q $(-\frac{1}{2},-5)$

7. Hello everyone

This is potentially a tricky question, as some of you have found. I agree with mathaddict's answer: Q is $(-\tfrac12, -5)$, 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 $(0, -3)$ is the mid-point of BC.

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

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 $\angle BAC = \theta$. Then, from the coordinates of A and B, and the fact that AC is vertical:
$\tan \theta = \frac{8}{6}=\frac43$.
And $\angle QAC = \tfrac12\theta \Rightarrow$ gradient of QA $= \cot\tfrac12\theta = m$, say.

Now $\tan\theta =\frac43= \frac{2\tan\tfrac12\theta}{1-\tan^2\tfrac12\theta}$
$\Rightarrow\frac43= \frac{2\left(\dfrac1m\right)}{1-\left(\dfrac1m\right)^2}$
$=\frac{2m}{m^2-1}$
$\Rightarrow 4m^2 -4 = 6m$

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

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

$\Rightarrow m = 2$ (the other root representing the gradient of the line perpendicular to this)
So the equation of AQ is
$y-4=2(x-4)$

i.e. $y = 2x-4$
This meets MQ where
$2x-4=4x-3$
i.e. at the point $(-\tfrac12, -5)$.