If the lines 3x + y - 4 = 0,x - alpha y + 10 | vedclass

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(B) The lines $3x + y - 4 = 0$ and $3x + y + k = 0$ are parallel,representing opposite sides of the square. Let the slopes be $m_1 = -3$ and $m_2 = -3

Vedclass · Accepted Answer

$-512$

Vedclass · Answer

(B) The lines $3x + y - 4 = 0$ and $3x + y + k = 0$ are parallel,representing opposite sides of the square. Let the slopes be $m_1 = -3$ and $m_2 = -3$. Since the sides are perpendicular,the slope of the other pair of sides must be $m_3 = \frac{1}{3}$. For the line $x - \alpha y + 10 = 0$,the slope is $\frac{1}{\alpha}$. Thus,$\frac{1}{\alpha} = \frac{1}{3} \Rightarrow \alpha = 3$. For the line $\beta x + 2y + 4 = 0$,the slope is $-\frac{\beta}{2}$. Thus,$-\frac{\beta}{2} = \frac{1}{3} \Rightarrow \beta = -\frac{2}{3}$. The distance between parallel lines $3x + y - 4 = 0$ and $3x + y + k = 0$ is $d = \frac{|k - (-4)|}{\sqrt{3^2 + 1^2}} = \frac{|k + 4|}{\sqrt{10}}$. The distance between parallel lines $x - 3y + 10 = 0$ and $-\frac{2}{3}x + 2y + 4 = 0$ (which is $x - 3y - 6 = 0$ after multiplying by $-\frac{3}{2}$) is $d = \frac{|10 - (-6)|}{\sqrt{1^2 + (-3)^2}} = \frac{16}{\sqrt{10}}$. Since it is a square,the distances must be equal: $\frac{|k + 4|}{\sqrt{10}} = \frac{16}{\sqrt{10}} \Rightarrow |k + 4| = 16$. Then,$\alpha \beta (k + 4)^2 = (3) \left(-\frac{2}{3}ight) (16)^2 = -2 	imes 256 = -512$.

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

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