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

Vertices of triangle are:
$A\equiv (a,0)$
$B\equiv (0,b)$
$C\equiv (0,0)$

Incentre:
$O_x=\frac{a\cdot A_x+b\cdot B_x+c\cdot C_x}{p}$

$O_y=\frac{a\cdot A_y+b\cdot B_y+c\cdot C_y}{p}$

Where:

$A_x,A_y$ are $x$ and $y$ coordinates of the point

$a,b,c$ are sides of the triangle

$p$ is perimeter.

$p=a+b+$$\sqrt{a^2+b^2}$

$\therefore$ incentre is $\left(\frac{ab}{a^{}+b^{}+\sqrt{a^2+b^2}}\right)$,$\left(\frac{ab}{a^{}+b^{}+\sqrt{a^2+b^2}}\right)$