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Question Number 54322 by peter frank last updated on 02/Feb/19

	Answered by kaivan.ahmadi last updated on 02/Feb/19

	$ax + by = c and bx - ay = d are perpendicular$ ${abx}^{2} + (b^{2} - a^{2}) xy - {aby}^{2} = cd$ $ab = 6$ $k = - ab = - 6$
	Answered by tanmay.chaudhury50@gmail.com last updated on 02/Feb/19

	${k y}^{2} - x y + 6 x^{2} = 0$ $y^{2} + (- \frac{1}{k}) x y + (\frac{6}{k}) x^{2} = 0$ $(y - m_{1} x) (y - m_{2} x) = 0$ $y^{2} - (m_{1} + m_{2}) x y + m_{1} m_{2} x^{2} = 0$ $- (m_{1} + m_{2}) x y = - \frac{1}{k} x y$ $m_{1} + m_{2} = \frac{1}{k}$ $m_{1} m_{2} = \frac{6}{k}$ $i) t w o ⊥ l i n e s m_{1} m_{2} = \frac{6}{k} = - 1 s o k = - 6$ $i i i) t a n \frac{π}{4} = \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} = \frac{\sqrt{{(m_{1} + m_{2})}^{2} - 4 m_{1} m_{2}}}{1 + m_{1} m_{2}}$ $1 = \frac{\sqrt{\frac{1}{k^{2}} - \frac{24}{k}}}{1 + \frac{6}{k}}$ $1 + \frac{12}{k} + \frac{36}{k^{2}} = \frac{1}{k^{2}} - \frac{24}{k}$ $k^{2} + 12 k + 36 = 1 - 24 k$ $k^{2} + 36 k + 35 = 0$ $(k + 1) (k + 35) = 0$ $k = - 1 a n d k = - 35$
	Commented by peter frank last updated on 02/Feb/19

	$thank you very much$
	Answered by mr W last updated on 02/Feb/19

	$6 x^{2} - x y + {k y}^{2} = (a x + b y) (c x + d y)$ $6 x^{2} - x y + {k y}^{2} = {a c x}^{2} + (a d + b c) x y + {b d y}^{2}$ $a c = 6$ $a d + b c = - 1$ $b d = k$ $s e t a = 1$ $\Rightarrow c = 6$ $d + 6 b = - 1$ $b d = k$ $b (6 b + 1) + k = 0$ $6 b^{2} + b + k = 0$ $\Rightarrow b = \frac{- 1 \pm \sqrt{1 - 24 k}}{12}$ $(i) :$ $\frac{a}{b} = - \frac{d}{c} \Rightarrow a c = - b d$ $\Rightarrow k = b d = - a c = - 6$ $(i i) :$ $D = 1 - 24 k ⩾ 0$ $\Rightarrow k ⩽ \frac{1}{24}$ $i . e . i f k > \frac{1}{24} \Rightarrow n o g r a p h$ $i f k = \frac{1}{24} \Rightarrow t h e g r a p h i s a s i n g l e l i n e$ $i f k < \frac{1}{24} \Rightarrow t h e g r a p h i s t w o l i n e s$ $(i i i) :$ $m_{1} = \frac{a}{b} = \tan θ_{1}$ $m_{2} = \frac{c}{d} = \tan θ_{2}$ $θ_{1} - θ_{2} = \pm \frac{π}{4}$ $\tan (θ_{1} - θ_{2}) = \pm 1$ $\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} = \pm 1$ $\frac{\frac{a}{b} - \frac{c}{d}}{1 + \frac{a c}{b d}} = \pm 1$ $\frac{a d - b c}{a c + b d} = \pm 1$ $\frac{1 + 2 b c}{6 + k} = \pm 1$ $\sqrt{1 - 24 k} = \pm (6 + k)$ $1 - 24 k = 36 + 12 k + k^{2}$ $35 + 36 k + k^{2} = 0$ $k = \frac{- 36 \pm \sqrt{36^{2} - 4 \times 35}}{2} = - 18 \pm 17$ $\Rightarrow k = - 1 o r - 35$
	Commented by mr W last updated on 02/Feb/19

	Commented by mr W last updated on 02/Feb/19

	Commented by mr W last updated on 02/Feb/19

	Commented by mr W last updated on 02/Feb/19

	Commented by peter frank last updated on 02/Feb/19

	$thank you very much sir .$
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