Question

A triangle ABC is formed by the points A(2,3),B (-2,-3),C (4,-3). what points of intersection of sides BC and bisector of angle A?

Answer 1

first find equation of line form by all points
now equation of line AB :-
(y-3)=(3+3)/(2+2)(x-3)
4 (y-3)=6 (x-2)
2y-6=3x-6
3x-2y=0
in the same way equation of lineBC:-
y+3=0
and equation of line AC:-
3x+y-9=0
now bisector of angle A,
(3x-2y)/root {13)=+_(3x+y-9)/root10

(3x-2y)/root13=(3x+y-9)/root10
root10 (3x-2y)=root13 (3x+y-13)--------(1)
now intersection of equation (1) and line BC
y=-3 and put this equation (1)
x=(6root10 +16root13)/3 (root13-root10)

in the same way ,
(3x-2y)/root13=-(3x+y-9)/root10
solve and another point

Answer 2

Solution :-

Let the point of intersection on BC be D

AB is the angle bisector of ∠A

By using angle bisector theorem :-

⇒ BD/DC = AB/AC ---eq(1)

Now find the lengths of AB and AC using distance formula

$d = \sqrt{(x_2 -x_1)^{2} + (y_2 - y_1)^{2} }$

A(2,3) B(-2,-3)

$\implies AB = \sqrt{( - 2 -2)^{2} + ( - 3 - 3)^{2} }$

$\implies AB = \sqrt{4^{2} + 6^{2} } = \sqrt{16 + 36} = \sqrt{52}$

$\implies AB = \sqrt{4 \times 13} = 2 \sqrt{13}$

A(2,3) C(4, - 3)

$\implies AC = \sqrt{(4 -2)^{2} + ( - 3 - 3)^{2} } =$

$\implies AC = \sqrt{2^{2} + 6^{2} } = \sqrt{4 + 36}$

$\implies AC = \sqrt{40} = \sqrt{4 \times 10} = 2 \sqrt{10}$

Substituting AB = 2√13 and BC = 2√10 in eq(1)

⇒ BD/DC = AB/AC

⇒ BD/DC = 2√13/2√10

⇒ BD/DC = √13/√10

⇒ BD : DC = √13 : √10

So, the line segment BC is divided in the ratio of √13 : √10

Now, by using section formula

$D(x,y) = \bigg( \dfrac{mx_2 + nx_1}{m + n} ,\dfrac{my_2 + ny_1}{m + n} \bigg)$

B(-2,-3) C(4,-3) m : n = √13 : √10

$\implies D(x,y) = \bigg( \dfrac{ \sqrt{13}(4)+ \sqrt{10}( - 2) }{ \sqrt{13} + \sqrt{10} } ,\dfrac{ \sqrt{13}( - 3) + \sqrt{10}( - 3) }{ \sqrt{13} + \sqrt{10} } \bigg)$

$\implies D(x,y) = \bigg( \dfrac{ 4\sqrt{13} - 2 \sqrt{10} }{ \sqrt{13} + \sqrt{10} } ,\dfrac{ - 3 \sqrt{13} - 3\sqrt{10} }{ \sqrt{13} + \sqrt{10} } \bigg)$

Substituting √13 = 3.606, √10 = 3.162

$\implies D(x,y) = \bigg( \dfrac{ 4(3.606)- 2 (3.162)}{ 3.606+ 3.162 } ,\dfrac{ - 3(3.606) - 3(3.162)}{ 3.606 + 3.162 } \bigg)$

$\implies D(x,y) = \bigg( \dfrac{14.424- 6.324}{ 3.606+ 3.162 } ,\dfrac{ - 10.818 - 9.486}{ 3.606 + 3.162 } \bigg)$

$\implies D(x,y) = \bigg( \dfrac{8.1}{6.768} ,\dfrac{ -20.304}{6.768 } \bigg)$

$\implies D(x,y) = (1.12 , - 3)$

Therefore the coordinates of point of intersection are (1.12, - 3)