Question

Here's a question to test your Mathematics !

Locus of Orthocentre of the variable triangle formed by (3,4), ( 5sinθ, 5cosθ), ( 5sinθ, -5cosθ) is ?

Answer 1

Let the points be

A ( 3, 4)

B ( 5 sinθ, 5 cosθ)

C ( 5sinθ, -5cosθ)

On finding distance between the points and Origin, You will find that The points are equidistant from origin.

Hence, Origin ( 0, 0) is the circumcentre of the given variable triangle.

We are aware of the relation that,

3G = 2S + H ( G - Centroid, S - Circumcentre, H - Orthocentre)

Here, Since S is (0, 0) . We can write 3G = H

Now,

3G = 3 + 5 sinθ + 5 sinθ, 4 + 5 cosθ - 5 cosθ

Let P(h, k) be any point on the locus.

h = 3 + 5 ( 2 sinθ), k = 4

h - 10 sinθ + 1 = 4, k = 4

So,

h - 10 sinθ + 1 = k

h - k = 10 sinθ - 1

Therefore, h - k + 1 - 10sinθ is the equation of the locus of the Orthocentre.

Answer 2

Step-by-step explanation:

Coordinates of ΔABC are A(3,4), B(5cosθ,5sinθ) and C(5sinθ,−5cosθ). Find the locus of its orthocenter.

My idea: It is clear that (0,0) is equidistant from the three coordinates. So S(0,0) is the circumcenter of ΔABC. Also the centroid of ΔABC is

G(5cosθ+5sinθ+33,5sinθ−5cosθ+43)

Now if the orthocenter is H(h,k), we have G divides H and S in the ratio 2:1. So G is given by

G(h3,k3)⟹hk=5cosθ+5sinθ+3=5sinθ−5cosθ+4

Eliminating θ we get the locus as:

(x−3)2+(y−4)2=50

Let me know whether my approach is logical .. if there is an alternate way I will be happy to know.