Question

the vertices of a triangle are (1,√3),(2cos teeta,2sin teeta) and (2sin teeta,-2 cos teeta) where teeta€R. the locus of ortho centre of the triangl​

Answer 1

Given:

The vertices of a triangle are (1,√3),(2cos teeta,2sin teeta) and (2sin teeta,-2 cos teeta) where teeta€R.  

To find:

The locus of orthocentre of the triangle​

Solution:

From given, we have,

The vertices of a triangle are (1,√3),(2cos teeta, 2sin teeta) and (2sin teeta,-2 cos teeta)

Let O(0, 0) be the circumceter

A (1,√3), B (2cos teeta, 2sin teeta) and C (2sin teeta,-2 cos teeta)

OA = √(1 + 3) =  2

OB = √[2² cos² ∅ + 2² sin² Ф] = 2

OC = √[(-2)² cos² ∅ + (-2)² sin² Ф] = 2

G = [(1+2 cos Ф + 2 sin Ф)/3 ,  (√3-2 cos Ф + 2 sin Ф)/3]

⇒ G = h/3, k/3

∴ h = 1+2 cos Ф + 2 sin Ф

⇒ h - 1 = 2 (cos Ф + sin Ф)  ...........(1)

∴ k = √3-2 cos Ф + 2 sin Ф

⇒ k - √3 = 2(sin Ф - cos Ф)  ..........(2)

squaring and adding equations (1) and (2), we get,

(h - 1)² + (k - √3)² = [2 (cos Ф + sin Ф)]² + [2(sin Ф - cos Ф)]²

(h - 1)² + (k - √3)² = 8

⇒ (x - 1)² + (y - √3)² = 8

∴ (x - 1)² + (y - √3)² = 8 is the locus of the orthocenter of triangle