Question

Find the equation of the bisector of angle A of a triangle whose vertices are A(4,3), B(0,0) and C(2,3)....



urgent.......

Answer 1

According to angle bisector theorem,

In a triangle say ABC,

The bisector of angle A will meet at BC say at a point called X

And hence AX divides BC in the ratio AC:AB

So according to the theorem AC:AB=CX:XB

Basically from A vertex to the other 2 vertices.

In the given triangle hence,

Lets find AC:AB

We will find by distance formula.

AC=$\sqrt{(x2-x1)^2+(y2-y1)^2}$

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

$\implies \sqrt{ (-2)^2+0}$

$\implies \sqrt{4}$

$\implies 2$

Now find AB:

AB=$\sqrt{(x2-x1)^2+(y2-y1)^2}$

$\implies \sqrt{(4-0)^2+(3-0)^2}$

$\implies \sqrt{16+9}$

$\implies \sqrt{25}$

$\implies 5$

Clearly from (1) and (2) we get:

AC:AB=CX:XB

==> 2:5=CX:CB

Now according to section formula:

Let X be (x,y)

Let the ratio in which BC is divided by X be m:n

So m= 2

n = 5

x1,y1=2,3

x2,y2=0,0

==> $x = \frac{mx2+nx1}{m+n}$

$\implies x = \frac{2*0+2*5}{2+5}$

$\implies x=\frac{10}{7}$

$\implies \frac{10}{7}$

$y = \frac{my2+ny1}{m+n}$

$\implies y=\frac{2*0+5*3}{5+2}$

$\implies y=\frac{15}{7}$

X=(10/7),(15/7)

Now calculate the slope of AX

x1,y1=4,3

x2,y2=(10/7,15/7)

$m=\frac{y2-y1}{x2-x1}$

$\implies \frac{15/7-3}{10/7-4}$

$\implies \frac{ (15-21)/7}{(10-28)/7}$

$\implies \frac{-6}{-18}$

$\implies \frac{1}{3}$

Now this is really tiring:

y-y1=m(x-x1)

Now A = (4,3)

==> x1=4

==>y1=3

y-3=1/3(x-4)

==> 3(y-3)=x-4

==>3y-9=x-4

==>3y-x-5=0

==>x-3y+5=0

Thats the equation:

x-3y+5=0

Hope it helps!

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Answer 2

Answer:

According to angle bisector theorem,

In a triangle say ABC,

The bisector of angle A will meet at BC say at a point called X

And hence AX divides BC in the ratio AC:AB

So according to the theorem AC:AB=CX:XB

Basically from A vertex to the other 2 vertices.

In the given triangle hence,

Lets find AC:AB

We will find by distance formula.

AC=

Now find AB:

AB=

Clearly from (1) and (2) we get:

AC:AB=CX:XB

==> 2:5=CX:CB

Now according to section formula:

Let X be (x,y)

Let the ratio in which BC is divided by X be m:n

So m= 2

n = 5

x1,y1=2,3

x2,y2=0,0

==>  

X=(10/7),(15/7)

Now calculate the slope of AX

x1,y1=4,3

x2,y2=(10/7,15/7)

Now this is really tiring:

y-y1=m(x-x1)

Now A = (4,3)

==> x1=4

==>y1=3

y-3=1/3(x-4)

==> 3(y-3)=x-4

==>3y-9=x-4

==>3y-x-5=0

==>x-3y+5=0

Thats the equation:

x-3y+5=0

Hope it helps!