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.
This should help a lot. You can construct the vectors u and v for the two orthocenters, and then just get the magnitude of |u-v|
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To solve this problem using coordinate geometry, let's assume that the right-angled triangle has vertices A, B, and C, with the right angle at vertex C. Let A be the origin (0,0), and let the coordinates of B and C be (a,0) and (0,b), respectively.
Since the right angle at C is bisected, the perpendicular bisector of the hypotenuse AB passes through the vertex C. Let's find the equation of the line passing through the midpoint of AB (M), which is (a/2, b/2), and the foot of the altitude from C to AB (P), which is (a, 0).
1. Find the slope of the line CM:
Slope of CM = (b/2 - 0) / (a/2 - 0) = b/a.
2. Find the equation of the line CM using the slope-intercept form:
y = mx + c, where m is the slope of CM and c is the y-intercept.
Since CM passes through point M, we substitute the coordinates of M into the equation:
b/2 = (b/a)*(a/2) + c.
Simplifying this equation, we get:
b/2 = b/2 + c,
which implies c = 0.
Thus, the equation of line CM is y = (b/a)x, or bx = ay.
Now, let's find the equation of the altitude CP. Since C is the origin, we can directly use the equation x = a for the altitude CP.
Next, we need to find the point of intersection of lines CM and CP to determine the orthocenter of the smaller triangle.
3. Solve the simultaneous equations of CM (bx = ay) and CP (x = a):
bx = ay,
x = a.
Substituting x=a from equation CP into equation CM, we get:
ba = ay,
Dividing both sides by a, we get:
b = y.
So, the point of intersection of CM and CP is (a, b).
Now that we have the coordinates of the orthocenter of the smaller triangle, which is also the orthocenter of the original triangle, we can find the distance between the orthocenters of the smaller triangles.
4. Finding the distance between the orthocenters:
The distance between two points P(x₁, y₁) and Q(x₂, y₂) can be calculated using the distance formula:
d = √((x₂ - x₁)² + (y₂ - y₁)²).
In this case, let the coordinates of the orthocenter be (x, y) = (a, b). Plugging in these values into the distance formula, we get:
d = √((a - 0)² + (b - 0)²),
d = √(a² + b²).
Therefore, the distance between the orthocenters of the smaller triangles is √(a² + b²).