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
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.
Given an isosceles triangle with vertices: .
Then point 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)
We know that: .
. . and we see that: .
So we have: .
This simplifies to a quadratic in
Quadratic Formula: .
. . and the positive root is: .