A(1 3) and c(-2/5 -2/5) are the two vertices of the triangle abc and equaion of angle bisector of abc is x+y=2 then,equation of side is
Answers
Given : A(1 3) and c(-2/5 -2/5) are the two vertices of the triangle abc and equation of angle bisector of abc is x+y=2
To find :
Solution:
Equation of angle bisector of ABC is x+y=2 => y = 2 - x
slope = - 1
Let say point B is x , ( 2- x)
Slope of AC = (3 - (-2/5)/(1 -(-2/5) = 17/7
y - 3 = (17/7)(x - 1)
=> 7y - 21 = 17x - 17
=> 17x - 7y = - 4
AC Equation = 17x - 7y = - 4
Slope of AB = (2 - x - 3)/(x - 1) = (-x - 1)/(x - 1) = (x + 1)/(1 - x)
Slope of BC = (2 - x - (-2/5))/(x - (-2/5))
= (10 - 5x + 2)/(5x + 2)
= (12 - 5x)/(5x + 2)
Slope of AB = (x + 1)/(1 - x)
Slope of angle bisector = -1
Slope of BC = (12 - 5x)/(5x + 2)
| {(x + 1)/(1 - x) - (-1) } /{1 + (-1)(x+ 1)/(1 - x)} | = | {(12 - 5x)/(5x+2) - (-1) } /{1 + (-1)(12 - 5x)/(5x+2)} |
=> | 2 / -2x | = | 14 / (10x - 10) |
=> | -1/x | = | 7/(5x - 5) |
-1/x = 7/(5x - 5)
=> -5x + 5 = 7x => x = 5/12 y = 2 - 5/12 = 19/12
( 5/12 , 19/12)
( 5/12 , 19/12) is intersection of x+y=2 and 17x - 7y = - 4 line AC
Hence discarded
1/x = 7/(5x - 5)
=> 5x - 5 = 7x => 2x = -5 => x = -5/2 , y = 2 -(-5/2) = 9/2
( - 5/2 , 9/2)
A(1 3) and c(-2/5 -2/5) B = ( - 5/2 , 9/2)
slope of AB = ( 9/2 - 3)/(-5/2 - 1) = -3/7
y - 3 = (-3/7)(x - 1)
=> 7y - 21 = -3x + 3
=> 3x + 7y = 24
Slope of BC = (9/2 + 2/5)/(-5/2 + 2/5) = -49/21 = -7/3
BC = y + 2/5 = (-7/3)(x + 2/5)
=> 3(5y + 2) = -7(5x + 2)
=> 35x + 15y = -20
=> 7x + 3y = - 4
AC Equation = 17x - 7y = - 4
AB Equation = 3x + 7y = 24
BC Equation 7x + 3y = - 4
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