# Math Help - geometry

1. ## geometry

plz help...

Suppose we are given an isosceles triangle ABC in the coordinate plane R^2 whose vertices are given by A = (o, h), B= (-x, 0), and C= (x, 0). Then the point J where the angle bisectors meet lies on the y-axis. Find its
y-coordinate.

I tried to approach this problem by taking the square of a, letting a = tan beta x. is that the correct way? Thank you so much for the help.

2. Originally Posted by jenjen
plz help...

Suppose we are given an isosceles triangle ABC in the coordinate plane R^2 whose vertices are given by A = (o, h), B= (-x, 0), and C= (x, 0). Then the point J where the angle bisectors meet lies on the y-axis. Find its
y-coordinate.

I tried to approach this problem by taking the square of a, letting a = tan beta x. is that the correct way? Thank you so much for the help.
Maybe. What's a?

I would do this as follows:

Find one of the base angles (I think you were calling this $\beta$) by
$h = x~tan(\beta)$

Then the y coordinate of the point of intersection of the two bisectors will be determined by
$y = x~tan \left ( \frac{\beta}{2} \right )$

-Dan

3. Hello, jenjen!

Given an isosceles triangle $ABC$ with vertices: . $A (0,h),\;B(-x,0),\;(x, 0)$
Then point $J$ where the angle bisectors meet lies on the y-axis.
Find its y-coordinate.
Code:
                A|
*(0,h)
/|\
/ | \
/  |  \
/   |   \
/    | J  \
/     *     \
/ θ *  |y     \
B / * θ   |       \ C
- - * - - - -+- - - - * - -
(-x,0)      D      (x,0)

Let $\theta = \angle JBC = \angle ABJ$

We know that: . $\tan2\theta \:=\:\frac{2\tan\theta}{1-\tan^2\!\theta} \:=\:\frac{h}{x}$

. . and we see that: . $\tan\theta \:=\:\frac{y}{x}$

So we have: . $\frac{2\tan\theta}{1-\tan^2\!\theta} \:=\:\frac{h}{x}\quad\Rightarrow\quad \frac{2\left(\frac{y}{x}\right)}{1-\left(\frac{y}{x}\right)^2} \:=\:\frac{h}{x}$

This simplifies to a quadratic in $y\!:\;\;hy^2 + 2x^2y - hx^2\:=\:0$

Quadratic Formula: . $y \;=\;\frac{-2x^2\pm\sqrt{4x^4 + 4h^2x^2}}{2h} \;=\;\frac{-x^2\pm x\sqrt{x^2+h^2}}{h}$

. . and the positive root is: . $y \;=\;\frac{x\sqrt{x^2+h^2} - x^2}{h}$