Straight Lines and Pair of Straight Lines

We have the following equations representing the sides of a square:$3x + y - 4 = 0$$x - \alpha y + 10 = 0$$\beta x + 2y + 4 = 0$$3x + y + k = 0$These lines form pairs of parallel lines: equations (i) with (iv) and (ii) with (iii).Equations (i) and (ii) are perpendicular, so their slopes must multiply to -1 :$(m_1 \times m_2) = -1$From equation (i), the slope $m_1 = -3$. From equation (ii), the slope $m_2 = \frac{1}{\alpha}$. Thus:$(-3) \times \frac{1}{\alpha} = -1$$\Rightarrow \alpha = 3$Similarly, equations (iii) and (iv) are perpendicular:$\left(\frac{-\beta}{2}\right) \times (-3) = -1$$\Rightarrow \beta = -\frac{2}{3}$Substituting $\alpha$ and $\beta$ back into the equations, we have:From equation (ii), $x - 3y + 10 = 0$.From equation (iii), $-\frac{2}{3}x + 2y + 4 = 0$, simplifying gives: $x - 3y - 6 = 0$Since the distances between the parallel sides must be equal for the square, we equate:$\frac{|k+4|}{\sqrt{3^2+1^2}} = \frac{|10+6|}{\sqrt{1^2+(-3)^2}}$Which simplifies to:$|k+4| = 16$Thus, the expression $\alpha \beta (k+4)^2$ becomes:$3 \times \left(-\frac{2}{3}\right) (16)^2 = -512$