A right angled triangle has sides a and b where a<b. If the right angle is bisected then then find the distance between orthocentres of the smaller triangles using coordinate geometry.
http://formulas.tutorvista.com/math/orthocenter-formula.html
This will give you the orthocenter, if you have the vertices of a triangle.
If you set up your right triangle with the right angle at C=(0,0) then let the other two vertices be at A=(a,0) and B=(0,b).
Then the right-angle bisector intersects the hypotenuse at D=(d,d) where d = ab/(a+b). (why?)
Now you have the vertices for your two smaller triangles. Find their orthocenters and then the distance between them.
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To solve this problem using coordinate geometry, let's assume that the right-angled triangle is placed in the coordinate plane such that the right angle is at the origin (0,0) and one side lies along the x-axis.
Let's assume that the two legs of the triangle are represented by the equations y = mx and y = -x/m, where m > 0.
The point where the right angle is bisected is the midpoint of the hypotenuse. Let's call this point M, which has coordinates P(a/2, b/2).
To find the orthocenter, we need to first find the equations of the altitudes of the smaller triangles formed by the bisecting line.
1. Triangle OAB:
- The altitude from O (vertex of the right angle) to AB can be represented by the equation y = -x/m.
- The altitude from A to OB can be represented by the equation y = mx - b.
2. Triangle OBC:
- The altitude from O to BC can be represented by the equation y = mx.
- The altitude from B to OC can be represented by the equation y = -x/m + b.
Now, let's find the coordinates of the orthocenters of the smaller triangles.
1. Orthocenter H1 of triangle OAB:
- Solve the equations of the altitudes y = -x/m and y = mx - b to find their point of intersection. Let's call this point H1(x1, y1).
- Solve the system of equations: -x/m = mx - b.
- Simplifying, we get: x = bm/(m^2 + 1) and y = -bx/m^2.
- Therefore, the coordinates of H1 are H1(bm/(m^2 + 1), -bm^2/(m^2 + 1)).
2. Orthocenter H2 of triangle OBC:
- Solve the equations of the altitudes y = mx and y = -x/m + b to find their point of intersection. Let's call this point H2(x2, y2).
- Solve the system of equations: mx = -x/m + b.
- Simplifying, we get: x = bm^2/(m^2 + 1) and y = bx/m^2 + b.
- Therefore, the coordinates of H2 are H2(bm^2/(m^2 + 1), bm^2/(m^2 + 1) + b).
To find the distance between the orthocenters H1 and H2, we can use the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2).
Substituting the coordinates of H1 and H2 into the distance formula, we get:
d = sqrt((bm^2/(m^2 + 1) - bm/(m^2 + 1))^2 + (-bm^2/(m^2 + 1) - (bm^2/(m^2 + 1) + b))^2).
Simplifying this expression will give you the distance between the orthocenters of the smaller triangles.