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JEE Main202226 Jul 2022Morning ShiftMathematicsStraight LinesActual

The equations of the sides A B , B C and C A of a triangle A B C are 2 x + y = 0 , x + p y = 15 a and x - y = 3 respectively. If its orthocentre is 2 , a , - 1 2 < a < 2 , then p is equal to

Correct answer

0

Step-by-step solution

Given the equations of the sides A B , B C and C A of a triangle A B C are 2 x + y = 0 , x + p y = 15 a and x - y = 3 respectively. Now on solving equation 2 x + y = 0 , x + p y = 15 a and x - y = 3 we get coordinates of A 1 , - 2 , B 15 a 1 - 2 p , - 30 a 1 - 2 p and C = 18 p - 30 p + 1 , 15 p - 33 p + 1 And let orthocentre be H 2 , a Slope of A H = a + 2 1 Sloe of B C = - 1 p As A H ⊥ B C , so p = a + 2 Now coordinate of C = 18 p - 30 p + 1 , 15 p - 33 p + 1 So, slope of H C = 15 p - 33 p + 1 - a 18 p - 30

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