Question

$A(a+1,a-1)$, $B(a^2+1,a^2-1)$ and $C(a^3+1,a^3-1)$ are given points. $D(11,9)$ is the mid point of $AB$ and $E(41,39)$ is the mid point of $BC$. If $F$ is the mid point of $AC,$ then $\frac{(BF{)}^2}{81}$ is equal to

Answer 1

Using mid-point formula,
$\frac{a+1+a^2+1}{2}=11,\frac{a-1+a^2-1}{2}=9$
$\Rightarrow a+a^2-20=0\Rightarrow a=-5$ or $4$
And $\frac{a^2+1+a^3+1}{2}=41,\frac{a^2-1+a^3-1}{2}=39$
$\Rightarrow a^2+a^3=80$ which holds for $a=4$
So, the given points are $A(5,3),B(17,15),C(65,33)$ and coordinates of $F$ are $(\frac{65+5}{2},\frac{63+3}{2})=(35,33)$ and
$(BF{)}^2=(35-17{)}^2+(33-15{)}^2$
$=(18{)}^2+(18{)}^2=648$

Thus the final answer is $\frac{\left(B{F}^2\right)}{81}=\frac{648}{81}=8$