Question

If incentre of triangle whose vertices are $(0,0),(4,0),(0,-3)$ is ($\alpha ,\beta$), then $\alpha ,\beta$ are roots of the equation .

• A. $x^2-x÷2=0$
• B. $x^2-1=0$
• C. $x^2-4x+3=0$
• D. $x^2+x-2=0$

Answer 1

Answer: (B)

$x^2-1=0$

Let $A=(0,0)$, $B=(4,0)$ and $C=(0,-3)$
The above triangle is a right angled triangle right angled at $A$.
Hence, $AB=4$, $BC=5$ and $AC=3$.
The co-ordinates of incentre are
$=(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})$
$=(\frac{AB\left(C_x\right)+BC\left(A_x\right)+AC\left(B_x\right)}{AB+BC+AC},\frac{AB\left(C_y\right)+BC\left(A_y\right)+AC\left(B_y\right)}{AB+BC+AC}$
$=(\frac{4\left(0\right)+5\left(0\right)+3\left(4\right)}{12},\frac{4(-3)+5\left(0\right)+3\left(0\right)}{12})$
$=(1,-1)$
Hence, $\alpha =1$ , $\beta =-1$
Therefore, $(x-1)(x+1)=0$
$x^2-1=0$ is the required equation.