Question

In $△ABC$ the coordinates of the vertex $A$ are $(4,-1)$ and lines $x-y-1=0$ and $2x-y=3$ are internal bisector of angles $B$ and $C$. Then, radius of incircle of triangle $△ABC$ is

• A. $\frac{4}{\sqrt{5}}$
• B. $\frac{3}{\sqrt{5}}$
• C. $\frac{6}{\sqrt{5}}$
• D. $\frac{7}{\sqrt{5}}$

Answer 1

The correct option is C $\frac{6}{\sqrt{5}}$

A] A (y,-4)

$x-y-1=0$ --(1)

$2x-y=3$ ---(2)

solving for (1), (2)

$x-y=1$ $2-y=1$

$2x-y=3$ $2-1=y$

- + - $\therefore$y=1

__________

$-3x=-2$

$\therefore x=2$

$(x,y)=(2,1)$

In centers is (2,1)

$1A=\sqrt{(4-2{)}^2+(-1-1{)}^2}$

$=\sqrt{4+4}$

$=\sqrt{8}$

$=2\sqrt{2}$

1A = r cosec $=\frac{A}{2}$

$r=\frac{A1}{cosesA/2}=\frac{2\sqrt{2}}{\sqrt{1+\frac{1}{9}}}=\frac{6}{\sqrt{5}}$