Question

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

The sides AB and AC of ΔABC are respectively 2x + 3y = 29 and x + 2y = 16 . If the mid point of BC is (5,6) then the equation of BC is

1] x + y - 10 = 0

2] x + y - 11 = 0

3] x + y - 9 = 0

4] x + y - 12 = 0

(Solve this question using concurrency of lines concept .Copied answers and spammers will be reported)​

Answer 1

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Answer 2

SOLUTION ::

Let Co-ordinate of B & C is (h, k) & ( p, q) respectively

Since the mid point of BC is (5,6)

So

( h+p) / 2 = 5 & ( k+q) /2 = 6

So p = 10-h & q = 12 - k

So Co-ordinate of C is ( 10-h, 12-k)

Now B(h, k) is a point on AB : 2x + 3y = 29

So 2h + 3k = 29 - - - - - - - (1)

Also

C( 10-h, 12-k) is a point on AC : x + 2y = 16

So 10 - h + 24 - 2k = 16

h + 2k = 18 - - - - - - - - - - - (2)

Solving Equation (1) & (2) we get

h = 4 , k = 7

So Co-ordinate of B is ( 4, 7 )

Co-ordinate of C is ( 10-4, 12-7 ) = ( 6, 5)

SO Equation of the line BC is given by

( y - 7)/( x-4) = ( 5-7)/(6-4)

=> ( y - 7)/( x-4) = - 2/2

=> x - 4 = - y+7

=> x + y - 11 = 0

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