Question

Three vertices of triangle ABC are A$(1,11)$, B$(9,8)$ and C$(15,2)$. The equation of angle bisector of angle A is

• A. $(8\sqrt{277}-14\sqrt{73})y+(3\sqrt{277}+9\sqrt{73})x-91\sqrt{277}+163\sqrt{73}=0$
• B. $(8\sqrt{277}-14\sqrt{73})y-(3\sqrt{277}-9\sqrt{73})x-91\sqrt{277}+163\sqrt{73}=0$
• C. $(8\sqrt{277}-14\sqrt{73})y+(3\sqrt{277}-9\sqrt{73})x-91\sqrt{277}+163\sqrt{73}=0$
• D. $(8\sqrt{277}+14\sqrt{73})y+(3\sqrt{277}-9\sqrt{73})x-91\sqrt{277}+163\sqrt{73}=0$

Answer 1

The correct option is C $(8\sqrt{277}-14\sqrt{73})y+(3\sqrt{277}-9\sqrt{73})x-91\sqrt{277}+163\sqrt{73}=0$

Given Points

$A(1,11),B(9,8),C(15,2)$

Equation of side AB

$y-11=\frac{8-11}{9-1}(x-1)$

$y-11=\frac{-3}{8}(x-1)$

$8(y-11)=-3(x-1)$

$8y-88=-3x+3$

$3x+8y-91=0$---(1)

Equation of side AC

$y-11=\frac{2-11}{15-1}(x-1)$

$y-11=\frac{-9}{14}(x-1)$

$14(y-11)=-9(x-1)$

$14y-154=-9x+9$

$9x+14y-163=0$---(2)

Eq of angle bisector of $AB,AC$

$\frac{3x+8y-91}{\sqrt{3^2+8^2}}=\pm \left(\frac{9x+14y-163}{\sqrt{9^2+{14}^2}}\right)$

$\frac{3x+8y-91}{\sqrt{9+64}}=\pm \left(\frac{9x+14y-163}{\sqrt{81+196}}\right)$

$\frac{3x+8y-91}{\sqrt{73}}=\pm \left(\frac{9x+14y-163}{\sqrt{277}}\right)$

On taking positive sign

$\frac{3x+8y-91}{\sqrt{73}}=\left(\frac{9x+14y-163}{\sqrt{277}}\right)$

$\sqrt{277}(3x+8y-91)=\sqrt{73}(9x+14y-163)$

$3\sqrt{277}x+8\sqrt{277}y-91\sqrt{277}=9\sqrt{73}x+14\sqrt{73}y-163\sqrt{73}$

$(8\sqrt{277}-14\sqrt{73})y+(3\sqrt{277}-9\sqrt{73})x-91\sqrt{277}+163\sqrt{73}=0$