Question

$\huge{\mathcal{\underline{\blue{Question:-}}}}$

If t₁ + t₂ + t₃ = -t₁t₂t₃ , Then orthocentre of the triangle formed by the points [ at₁t₂ , a(t₁+t₂) ] , [at₂t₃ , a(t₂+t₃) ] , [at₃t₁ , a(t₃+t₁) ] lies on

1] (a,0)

2] (-a,0)

3] (0,a)

4] (0,-a)

(Question related to concurrency of lines topic in coordinate geometry .Copied answers and spammers will be reported)

​

Answer 1

Answer:

Let ABC be the triangle formed by the points A [ at₁t₂ , a(t₁+t₂) ] , B [at₂t₃ , a(t₂+t₃) ] , C [at₃t₁ , a(t₃+t₁) ]

NOW slope of BC

= [a(t₂+t₃) - a(t₃+t₁) ] / [at₂t₃ - at₃t₁ ]

= [a(t₂ - t₁) ] / [t₃ a(t₂ - t₁) ]

= 1/t₃

Similarly slope of AC = 1/t₁

Therefore the equation of the line through A perpendicular to BC is

[ y - a(t₁+t₂) ] = - t₃ [ x - at₁t₂ ] - - - - - - - - (1)

Again the equation of the line through B perpendicular to AC is

[ y - a(t₂+t₃) ] = - t₁ [ x - at₂t₃ ] - - - - - - - (2)

Now Equation (1) - Equation (2) gives

[ y - a(t₁+t₂) ] - [ y - a(t₂+t₃) ] = - t₃ [ x - at₁t₂ ] + t₁ [ x - at₂t₃ ]

- a(t₁+t₂) + a(t₂+t₃) ] = - x t₃ - at₁t₂ t₃ + x t₁ + at₁t₂ t₃

a(t₃ - t₁) = - x(t₃ - t₁)

x = a

From Equation (1)

[ y - a(t₁+t₂) ] = - t₃ [ a - at₁t₂ ]

[ y - a(t₁+t₂) ] = - at₃ + at₁t₂ t₃

y = a ( t₁ + t₂ + t₃ + t₁t₂t₃) = 0

Now The point of intersection of Equation (1) and Equation (2) is the required orthocentre

So the required orthocentre is (a, 0)

Please Mark it Brainlist

Answer 2

Answer:

$\bold{\mathfrak{\boxed{\boxed{\green{DON'T \: REPORT\: PLEASE}}}}}$