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 ?
slope of AB =
slope of AC =
let the bisector of angle BAC intersects with BC at D.
hence slope of AD =
Equation of AD:
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,
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 ?
Originally Posted by ukorov
Oh....I read it wrong....C is (4, -4) and slope of AC is undefined (Angry)
Originally Posted by mathaddict
let me think...
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 clockwise as shown in attached figure.
slope of AB becomes -
slope of AX (AC) becomes 0
now the slope of AD can be calculated as (correct?)
therefore the slope of AD before rotation should be 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
hope there is no mistake this time(Worried)
Hi ukorov , this is what i got .
Given that BQ=QC
eventually simplify to ---- 1
Draw 2 perpendiculars QR and QS which are equal because
Then use the perpendicular distance formula
Solve the simultaneous equation , i got Q
This is potentially a tricky question, as some of you have found. I agree with mathaddict's answer: Q is , 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 is the mid-point of BC.
Now the gradient of BC gradient of MQ . So the equation of MQ iswhich 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 . Then, from the coordinates of A and B, and the fact that AC is vertical:
.And gradient of QA , say.
So the equation of AQ is
(the other root representing the gradient of the line perpendicular to this)
This meets MQ wherei.e. at the point .