Question

The vertices of a variable triangle are (3,4),(5cos theta,5sin theta), and (5sin theta,-5cos theta), where theta € R. The locus of its orthocentre is
A)(x+y-1)^2 +(x-y-7)^2=100
B)(x+y-7)^2 +(x-y-1)^2=100
C)(x+y-7)^2 +(x-y-1)^2=100
D)(x+y-7)^2+(x-y+1)^2=100​

Answer 1

Topic :-

Locus

Given :-

The vertices of a variable triangle are (3, 4), (5cosθ, 5sinθ) and (5sinθ, -5cosθ), where θ ∈ R.

To Find :-

The locus of its orthocenter.

Concept to be Used :-

Slope of a line pâssing through two points (m)

$\sf{m=\dfrac{y_2-y_1}{x_2-x_1}}$

Angle between two lines  

If θ be the angle between two lines, y = m₁x + c and y = m₂x + c, then

$\sf{\tan\theta=\pm\left(\dfrac{m_1-m_2}{1+m_1m_2} \right)}$

While calculating interior angles of a triangle, its interior angles are given by taking positive sign in the formula.

Coordinates of Orthocenter  

If vertices of the triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃) then coordinates of orthocenter :

$\sf{\left( \dfrac{x_1\tan A+x_2\tan B+x_3\tan C}{\tan A +\tan B + \tan C},\dfrac{y_1\tan A+y_2\tan B+y_3\tan C}{\tan A +\tan B + \tan C}\right)}$

Solution :-

There are many ways to solve this sum, we may calculate all required values which need to calculate orthocenter through its formula and then substitute them. This method will be lengthy and it will require lots of calculation.

Second method, we can calculate coordinates of Centroid and Circumcenter and then with the help of these we may calculate coordinates of orthocenter very easily. This method will be again a very length one.

Third method, since options are given we may take any particular case and then solve it using formula of coordinates of orthocenter. The option which will get satisfied by the point would be our required answer. Since, the option will be satisfied by orthocenter obtained through particular case, it will be also satisfied for general case. This method would be easy one.

We will be solving using third method.

Taking particular case θ = 0°,

Now, vertices of triangle are :-

A (x₁, y₁) ≡ A (3, 4)

B (x₂, y₂) ≡ B (5sinθ, -5cosθ) ≡ B (5sin(0°), -5cos(0°)) ≡ B (0, -5)

C (x₃, y₃) ≡ C (5cosθ, 5sinθ) ≡ C (5cos(0°), 5sin(0°)) ≡ B (5, 0)

$\sf{Slope\:of\:AB\:(m_1)=\dfrac{4-(-5)}{3-0}=\dfrac{4+5}{3}=3}$

$\sf{Slope\:of\:BC\:(m_2)=\dfrac{0-(-5)}{5-0}=\dfrac{5}{5}=1}$

$\sf{Slope\:of\:AC\:(m_3)=\dfrac{0-4}{5-3}=\dfrac{-4}{2}=-2}$

$\sf{\tan A=\dfrac{m_3-m_1}{1+m_3m_1}=\dfrac{-2-3}{1+(-2)(3)}=1}$

$\sf{\tan B=\dfrac{m_1-m_2}{1+m_1m_2}=\dfrac{3-1}{1+(3)(1)}=0.50}$

$\sf{\tan C=\dfrac{m_2-m_3}{1+m_2m_3}=\dfrac{1-(-2)}{1+(1)(-2)}=-3}$

Applying formula of coordinates of Orthocenter,

$\sf{\left( \dfrac{3(1)+0(0.5)+5(-3)}{1 + 0.5-3},\dfrac{4(1)-5(0.5)+0(-3)}{1 + 0.5-3}\right)}$

$\sf{\left( \dfrac{3-15}{-1.5},\dfrac{4-2.5}{-1.5}\right)}$

$\sf{\left( \dfrac{-12}{-1.5},\dfrac{1.5}{-1.5}\right)}$

$\sf{\left( 8,-1\right)\longrightarrow Orthocenter\:for\:particular\:case}$

Now, the option which will satisfy (8, -1) will be the required answer,

Checking option D,

(x + y - 7)² + (x - y + 1)²

Substituting values,

(8 - 1 - 7)² + (8 - (-1) + 1)²  

0² + 10²

100  

LHS = RHS

Hence, option D satisfies obtained point.

Note : As this option is satisfied by a particular case of θ, it will be also satisfied by the general value of θ.

Answer :-

The locus of orthocenter of given variable triangle is (x + y - 7)² + (x - y + 1)² = 100.

Hence, option D is correct option.

Note : There are many more possible methods to solve this sum except mentioned three methods.