Question

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

• A. x-3y+5=0
• B. x+3y-5=0
• C. 3x-y-9=0
• D. 3x+y-9=0

Answer 1

The correct option is A x-3y+5=0

In $\Delta$ ABC
$\mathrm{AB}=\sqrt{4^2+3^2}=5\mathrm{and}\mathrm{AC}=\sqrt{(4-2{)}^2+(3-3{)}^2}=2$
As AD bisects $\angle$A
$⟹$ AB:AC=BD:DC
$⟹$ BD:DC=5:2
Then, the coordinates of D will be
$(\frac{5\times 2+0}{5+2},\frac{5\times 3+0}{5+2})=(\frac{10}{7},\frac{15}{7})$
Now, slope of the line joining A and D is
$\frac{3-\frac{15}{7}}{4-\frac{10}{7}}=\frac{21-15}{28-10}=\frac{1}{3}$
$\therefore \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{joining}A\mathrm{and}D\mathrm{using}\mathrm{two}-\mathrm{point}\mathrm{form}\mathrm{will}\mathrm{be}(y-3)=\frac{1}{3}(x-4)\Rightarrow 3y-9=x-4\Rightarrow x-3y+5=0$