Question

The incentre of the triangle formed by the axes and the line $\frac{x}{a}+\frac{y}{b}=1$ is

• A. $(\frac{a}{2},\frac{b}{2})$
• B. $(\frac{ab}{a+b+\sqrt{ab}},\frac{ab}{a+b+\sqrt{ab}})$
• C. $(\frac{a}{3},\frac{b}{3})$
• D. $(\frac{ab}{a+b+\sqrt{a^2+b^2}},\frac{ab}{a+b+\sqrt{a^2+b^2}})$

Answer 1

The correct option is D $(\frac{ab}{a+b+\sqrt{a^2+b^2}},\frac{ab}{a+b+\sqrt{a^2+b^2}})$

The equation if in-circle is

$\Rightarrow$ $(x-r{)}^2+(y-r{)}^2=r^2$ ------ ( 1 )

The given line is $\frac{x}{a}+\frac{y}{b}=1$

$\Rightarrow$ $bx+ay=ab$

$\Rightarrow$ $b(x-r)+a(y-r)=ab-ar-br$

$\Rightarrow$ $y-r=\frac{-b}{a}(x-r)+\frac{ab-ar-br}{a}$ ------ ( 2 )

Now the line ( 2 ) will be tangent at $P$

The condition of tang-ency is

${\left(\frac{ab-ar-br}{a}\right)}^2=(-r{)}^2[{\left(\frac{-b}{a}\right)}^2+1]$ [ Since, $c^2=a^2(m^2+1)$ ]

$\Rightarrow$ $(ab-ar-br{)}^2=r^2(b^2+a^2)$

$\Rightarrow$ $(ab-ar-br)=r^2(b^2+a^2)$

$\Rightarrow$ $ab-ar-br=r\sqrt{(b^2+a^2)}$

$\Rightarrow$ $ab=r(a+b+\sqrt{a^2+b^2})$

$\Rightarrow$ $r=\frac{ab}{a+b+\sqrt{a^2+b^2}}$

$\therefore$ Incenter $=(\frac{ab}{a+b+\sqrt{a{a}^2+b^2}},\frac{ab}{a+b+\sqrt{a^2+b^2}})$