(Q2)A triangle ABC is formed by the points A(2,3) , B(-2,-3) C(4,-3).What is the point of intersection of side BC and angular bisector of angle A ?

Given △ABC , where A(2,3) , B(-2,-3) , C(4,-3)

Let AD be the bisector to angle A meeting BC at D

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

[The bisector of vertical angle of triangle divides the base in thr ratio of other two sides]

Distance formula = √((x2 - x1)2 + (y2 - y1)2)

AB = √((- - )2 + (- - )2)

  = √( + ) = √

AC = √(( - )2 + (- - )2)

  = √( + ) = √

AB:AC = √
= √
[∴ Square root for both Numerator and Denominator]

   = √ : √

Now D is a point which divides BC in the ratio √13:√10 internally section formula (x,y)

= (
m1x2+m2x1
m1+m2
 ,
m1y2+m2y1
m1+m2
 )
= (
√ ×4 + √10(-)
√ + √
,
√ ×(-3) + √(-3)
√ + √
)
= (
4√ - 2 √
√ + √
,
-3(√ + √)
√ + √
)

= ( - 2 √, -3)

(By rationalising denominator of the x-coordinate)