In a triangle ABC, if A (2, – 1) and 7x – 10y + 1 = 0 and 3x – 2y + 5 = 0 are equations of an altitude and an angle bisector respectively drawn from B, then equation of BC is
Answers
Given : triangle ABC A (2, – 1) equations of an altitude : 7x – 10y + 1 = 0 and equations of angle bisector 3x – 2y + 5 = 0 drawn from B
To find : Equation of BC
Solution:
A (2, – 1)
7x – 10y + 1 = 0 and 3x – 2y + 5 = 0 are equations of an altitude and an angle bisector respectively drawn from B
=> intersection is point B
=> 7x - 10y + 1 = 0
& 15x - 10y + 25 = 0
=> 8x + 24 = 0
=> x = - 3 & y = -2
=> B = ( - 3 , - 2)
A (2, – 1) B = ( - 3 , - 2)
Slope of AB = ( - 1 + 2)/(2 + 3) = 1/5
BX - angle bisector 3x – 2y + 5 = 0
=> y = 3x/2 + 5/2
slope = 3/2
Let Slope of BC = m
BX - angle bisector
=> | (m - (3/2) ) / ( 1 + m(3/2) | = | (1/5 - 3/2) / ( 1 + (1/5)(3/2) |
=> | (2m - 3) / (2 + 3m) | = | ( 2 - 15) / ( 10 + 3 |
=> | (2m - 3) / (2 + 3m) | = | - 1 |
=> | (2m - 3) / (2 + 3m) | = 1
=> (2m - 3) / (2 + 3m) = ± 1
2m - 3 = 2 + 3m => m = - 5
or 2m - 3 = - 2 - 3m => 5m = 1 => m = 1/5 ( slope of AB) ( hence not possible)
m = -5
y = -5x + c
passes through B (-3 , - 2)
=> -2 = 15 + c
=> c = -17
=> y = -5x -17
=> 5x + y + 17 = 0
equation of BC is 5x + y + 17 = 0
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