If the lines $3 x+y-4=0, x-\alpha y+10=0, \beta x+2 y+4=$ 0 and $3 x+y+k=0$ represent the sides of a square, then $\alpha \beta(k+4)^2=$

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Aicharya · Accepted Answer

Since, $m_{\mathrm{AB}}=-3, m_{\mathrm{BC}}=\frac{1}{\alpha}$ $\because m_{\mathrm{AB}} \cdot m_{\mathrm{BC}}=-1 \Rightarrow \frac{-3}{\alpha}=-1 \Rightarrow \alpha=3$ $\begin{aligned} & \text { and } m_{\mathrm{AD}}=\frac{-\beta}{2} \quad \because m_{\mathrm{AD}} \cdot m_{\mathrm{AB}}=-1 \Rightarrow(-3) \cdot\left(\frac{-\beta}{2}\right)=-1 \\ & \Rightarrow \beta=\frac{-2}{3}\end{aligned}$ Now, $\mathrm{AB}=\frac{|10+4|}{\sqrt{1^2+(-3)^2}}=\frac{16}{\sqrt{10}}$ and $\mathrm{BC}=\frac{|k+4|}{\sqrt{1+9}}=\frac{k+4}{\sqrt{10}}$ Since, $\mathrm{AB}=\mathrm{BC} \Rightarrow \frac{16}{\sqrt{10}}=\frac{k+4}{\sqrt{10}} \Rightarrow k=12$ Now, $\alpha \beta(k+4)^2=3 \times\left(\frac{-2}{3}\right) \times 256=-512$

If the lines $3 x+y-4=0, x-\alpha y+10=0, \beta x+2 y+4=$ 0 and $3 x+y+k=0$ represent the sides of a square,…

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