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

Let, $z_1$, $z_2$ be $(a,0)(b,0)$ respectively

Equation of $L_1=y-0=\frac{-1}{1}(x-3)$

$⟹y=-x+\mathrm{3......}\left(i\right)$

Equation of $L_2=y-0=\frac{1}{2}(x-0)$

$⟹y=\frac{1}{2}x$

$⟹2y=x....\left(ii\right)$

Since, $z_3(b,c)$ and $z_4(a,c)$ lies in eqn $L_1$ and eqn $L_2$respectively, substituting the values ,we get

$c+b=\mathrm{3....}\left(iii\right)$ and $2c=a....\left(iv\right)$

Solving eqn (iii) and eqn (iv) simultaneously, we get, $a+2b=6$

Now, Area of square =$(side{)}^2$

$=c^2=(b-a{)}^2$

Given: Area of triangle formed by $z_3,z_4$ and origin= $\frac{m}{n}$

$\therefore$ Area of square= $\frac{2m}{n}$ {$\because$ Area of triangle is half the area of parallellogram when on the same base and between same parallels}

Now, $c^2=\frac{2m}{n}$

$⟹(\frac{a}{2}{)}^2=\frac{2m}{n}$....from eqn (iv)

$⟹\frac{a^2}{8}=\frac{m}{n}....\left(v\right)$

and,

$(a-b{)}^2=\frac{2m}{n}$

$⟹(a-3+\frac{a}{2}{)}^2=\frac{2m}{n}$.....from eqn (iii) and (iv)

$⟹(\frac{3a}{2}-3{)}^2=\frac{2m}{n}....(vi)$

Equating eqn (v) and (vi), we get,

$⟹8a^2-36a+36=0$

$⟹(2a-3)(a-3)=0$

$\therefore a=3or\frac{3}{2}$

$\because a=3$ is not possible, $a=\frac{3}{2}$ is considered

Substituting $a=\frac{3}{2}$ in eqn (v), we get

Now, $\frac{(3/2{)}^2}{8}=\frac{m}{n}$

$\frac{m}{n}=\frac{9}{32}$

$\because$ m and n are in lowest form,

$m+n=9+32=41$