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The vertices of a triangle are A(-1,-1),B(3,5),C(-4,1). The internal and external angle bisectors of the triangle through A cut the side BC at E,F respectively. Then the true statements among the following are The abscissa of E is -5/3 The ordinate of E is -3 The abscissa of F is -11 The ordinate of F is 7/3

  1. The vertices of a triangle are A(-1,-1),B(3,5),C(-4,1). The internal and external angle bisectors of the triangle through A cut the side BC at E,F respectively. Then the true statements among the following are
  1. The abscissa  of E is -5/3  
  2.   The ordinate of E is -3 
  3.    The abscissa  of F is -11     
  4.   The ordinate of F is 7/3

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2 Answers

Samyak Jain
333 Points
4 years ago
A(– 1, – 1) , B(3,5) , C(– 4,1)
Equation of line through A and B is y + 1 = {(5+1)/(3+1)} (x + 1) after solving, we get 3x – 2y + 1 = 0. 
(two point form)
Equation of line through A and C is y + 1 = {(1+1)/(-4+1)} (x + 1) after solving, we get 2x + 3y + 5 = 0.
Equations of angle bisectors of lines through A and B & A and C are 
3x – 2y + 1  =  \pm (2x + 3y + 5)   i.e.  3x – 2y + 1 = 2x + 3y + 5   and   3x – 2y + 1 = – (2x + 3y + 5)
x – 5y – 4 = 0  and  5x + y + 6 = 0.
 
These lines are perpendicular. So, a good idea to distinguish internal and external bisectors is to roughly plot
A, B, C and observe that x – 5y – 4 = 0 is the equation of external bisector and 
Samyak Jain
333 Points
4 years ago
5x + y + 6 = 0 is the equation of internal bisector.
Equation of line through B and C using two-point form is 4x – 7y + 23 = 0.
E is the point of intersection of 5x + y + 6 = 0 and 4x – 7y + 23 = 0 while F is the point of intersection of
x – 5y – 6 = 0 and 4x – 7y + 23 = 0.
You will get E(– 5/3 , 7/3)  and  F(– 11 , – 3).
So, options a and c are correct.

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