1.

Find the equation of the internal bisector of angle `B A C`of the triangle `A B C`whose vertices `A ,B ,C`are`(5,2),(2,3)a n d(6,5)`respectively.

Answer» `(x_1,y_1),(x_2,y_2)`
`y-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)`
Equation of AC
`y-2=(5-2)/(6-5)(x-5)`
`y-2=3(x-5)`
`y-2=3x-15`
`3x-y-13=0`
Equation of AB
`y-2=(3-2)/(2-5)(x-5)`
`y-2=-1/3(x-5)`
`x+3y-11=0`
`|(a_1x+b_1y+c_1)/sqrt(a_1^2+b_1^2)|=pm|(a_2x+b_2y+c_2)/sqrt(a_2^2+b_2^2)|`
`3x-y-13=0`
`x+3y-11=0`
`(3x-y-13)/sqrt(9+1)=pmm(x+3y-11)/sqrt(1+9`
`3x-y-13=x+3y-11`
`2x-4y-2=0`
`3x-y-13=-x-3y+7`
`4x+2y-24=0`
`m=(-4)/2=-2`
`AB=x+3y-11=0`
`tantheta=|(m_1-m_2)/(1+m_1m_2)|=|((-1/3)-(-2))/(1+(-1/3)(-2))|`
`tantheta=((-1/3)+2)/(1+(2/3))`
`=(5/3)/(5/3)=1`
`tantheta=45^@`.


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