If $\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2=a^2,\left(x_2-x_3\right)^2+\left(y_2-y_3\right)^2=

Question

If $\left(x_1-x_2ight)^2+\left(y_1-y_2ight)^2=a^2,\left(x_2-x_3ight)^2+\left(y_2-y_3ight)^2=b^2$,
$\left(x_3-x_1ight)^2+\left(y_3-y_1ight)^2=c^2$ and $4\left|\begin{array}{lll}x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1\end{array}ight|$
= $\lambda\left\{\lambda^3-\left(\lambda_1+\lambda_2+\lambda_3ight) \lambda^2+\left(\lambda_1 \lambda_2+\lambda_2 \lambda_3+\lambda_3 \lambda_1ight) \lambda ight.$ $\left.-\lambda_1 \lambda_2 \lambda_3ight\}$ then

Professor PC Thomas Classes · Answer

Area of $	riangle P Q R=\frac{1}{2}\left|\begin{array}{lll}x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1\end{array}ight|$
$=\sqrt{\frac{(a+b+c)}{2} \frac{(a+b-c)}{2} \frac{(b+c-a)}{2} \frac{(c+a-b)}{2}}$
$4\left|\begin{array}{lll}x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1\end{array}ight|=(a+b+c)(b+c-a)(c+a-b)(a+b-c)$
 i.e $a+b+c=\lambda, 2 a=\lambda_1, 2 b=\lambda_2, 2 c=\lambda_3$
$\begin{aligned} & 	herefore 2 \lambda=\lambda_1+\lambda_2+\lambda_3 \ & \sum \lambda_1 \lambda_2=4(a b+b c+c a)=4 \sum a b \ & \lambda_1 \lambda_2 \lambda_3=8 a b c 	ext { and } \frac{a+b+c}{3} \geq(a b c)^{1 / 3}\end{aligned}$
$\Rightarrow \frac{\lambda}{3} \geq\left(\frac{\lambda_1}{2} \cdot \frac{\lambda_2}{2} \cdot \frac{\lambda_3}{2}ight)^{1 / 3} \Rightarrow \lambda \geq \frac{3}{2}\left(\lambda_1 \lambda_2 \lambda_3ight)^{1 / 3}$