Let $A B C D$ (taken in order) be a square. The coordinates of $A$ and $C$ are $(1, 3)$ and $(5, 1)$ respectively. Then the product of abscissae of $B$ and $D$ is

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Solution

Slope of $A C$ is $- \frac{1}{2}$
Let the equation of $B D$ with slope $2$ be $y = 2 x + c$
Middle point of $A C$ i.e., $(3, 2)$ lies on line.
Hence, $2 = 6 + c$
$\Rightarrow c = - 4$
Equation of $B D$ is $y = 2 x - 4$

Now, $A C = \sqrt{(5 - 1)^{2} + (1 - 3)^{2}} = 2 \sqrt{5}$
So, $B D = 2 \sqrt{5}$
$∴ M D = M B = \sqrt{5}$
Taking parametric form of equation of line $B D,$
$\frac{x - 3}{cos θ} = \frac{y - 2}{sin θ} = \pm \sqrt{5}$
$x = 3 \pm \sqrt{5} cos θ$
For line $B D,$
$tan θ = 2$
$\Rightarrow cos θ = \frac{1}{\sqrt{5}}$
So, $x = 3 \pm (\sqrt{5} \times \frac{1}{\sqrt{5}})$
$\Rightarrow x = 2$ or $x = 4$
Here, $2$ is the abscissa of point $B$ and $4$ is the abscissa of point $D$ .
Product of abscissae of $B$ and $D$ is $2 \times 4 = 8$

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Let ABCD (taken in order) be a square. The coordinates of A and C are (1,3) and (5,1) respectively. Then the product of abscissae of B and D is

Let $A B C D$ (taken in order) be a square. The coordinates of $A$ and $C$ are $(1, 3)$ and $(5, 1)$ respectively. Then the product of abscissae of $B$ and $D$ is