# ABOUT THE COORDINATE GEOMETRYThe vertices of triangle ABC are A(2,1), B(6,-2), C(8,9). If AD is angle bisector, where D on BC, then coordinates of D are A) (20/3,5/3) B) (5,8) ...

ABOUT THE COORDINATE GEOMETRY

The vertices of triangle ABC are A(2,1), B(6,-2), C(8,9). If AD is angle bisector, where D on BC, then coordinates of D are

A) (20/3,5/3) B) (5,8) C) (0,0) D) (14,3) E) NONE OF THESE.

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*The vertices of triangle ABC are A(2,1), B(6,-2), C(8,9). If AD is an angle bisector, with D on BC, then the coordinates of D are*

*A) (20/3,5/3) B) (5,8) C) (0,0) D) (14,3) E) NONE OF THESE.*

Using the distance formula (the distance from `(x_1,y_1)` to `(x_2,y_2)` is

`d=sqrt((x_1-x_2)^2+(y_1-y_2)^2)` ), we find AB=5, AC=10, and BC=`5sqrt(5)` .

Now if AD is an angle bisector, it divides the side opposite the angle proportionally to the other two sides. Thus `10/(CD)=5/(BD)` , or CD=2*BD.

Thus D divides BC into two segments in the ratio of 1:2, or BD=1/3(BC) and CD=2/3(BC). Then BD=`(5sqrt(5))/3` and CD=`(10sqrt(5))/3` .

Note that the line connecting B and C is given by `y=11/2x-35` . Then point D can be described as `(x,11/2x-35)` .

Now we apply the distance function to DC getting `(x-8)^2+(11/2x-35-9)^2=500/9` and the to DB getting

`(x-6)^2+(11/2x-35+2)^2=125/9`

`(x-8)^2+(11/2x-44)^2-500/9=(x-6)^2+(11/2x-33)^2-125/9`

Expanding the binomials and collecting like terms we get 125x=833 1/3 or x=20/3.

**Thus the answer is A: (20/3,5/3)**

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