Question

Let $z_i=(i=1,,2,3,4)$ represent the vertices of a square all of which lie on the sides of the triangle with vertices $(0,0)$, $(2,1)$ and $(3,0)$. If $z_1$ and $z_2$ are purely real, then area of triangle formed by $z_3$ $z_4$ and origin is $\frac{m}{n}$ (where $m$ and $n$ are in their lowest form). Find the value of $(m+n)$.

Answer 1

Since, $z_1$ and $z_2$ are purely real, let, $z_1=a_1$ and $z_2=a_2$ are represented by the points D and E on the real axis.

According to the problem, DEFG is a square so co-ordinates of F and G are $(a_2,a_2-a_1)$ and $(a_1,a_2-a_1)$ respectively.

Now, equation of OA is $\frac{y-1}{0-1}=\frac{x-2}{0-2}=$ $\Rightarrow x=2y$.....(1).

And G lies on OA then,

$a_1=2(a_2-a_1)$

or, $2a_2=3a_1$......(2).

Now, equation of AB is $\frac{y-0}{1-0}=\frac{x-3}{2-3}$ $\Rightarrow x+y=3$......(3).

Also, F lies on AB, then

$a_2+a_2-a_1=3$

or, $2a_2-a_1=3$......(4).

Using (2) in (4) we get, $a_1=\frac{3}{2}$ and $a_2=\frac{9}{4}$.

$\therefore$ co-ordinates of F and G are $(\frac{3}{2},\frac{3}{4})$ and $(\frac{9}{4},\frac{3}{4})$.

Now, area of OFG is $\frac{1}{2}\times |0(\frac{3}{2}-\frac{3}{2})+\frac{3}{2}(\frac{3}{4}-0)+\frac{9}{4}(0-\frac{3}{4})|=\frac{9}{32}$.

By the problem, $\frac{m}{n}=\frac{9}{32}$ $\Rightarrow (m+n)=(9+32)=41$.