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Question Number 140076 by bobhans last updated on 04/May/21

Commented by mr W last updated on 05/May/21

$e q n . 2 x - 3 y + a = 0 f i x e s o n l y t h e$ $i n c l i n a t i o n o f t h e l i n e . t h e v e r t i c a l$ $p o s i t i o n w i l l b e d e t e r m i n e d b y t h e$ $v a l u e o f a . t h e d i s t a n c e f r o m p o i n t$ $(3, - 6) m u s t t o t h i s l i n e m u s t b e t h e$ $s a m e a s t o 2 x + 3 y + 25 = 0, s o w e g e t$ $t w o p a r a l l e l l i n e s w i t h d i f f e r e n t a .$
Commented by EDWIN88 last updated on 04/May/21
$clockwise for θ (((x′−3)),((y^′ +6)) ) = ((( cos θ sin θ)),((−sin θ cos θ)) ) (((x−3)),((y+6)) ) (((x′−3)),((y′+6)) ) = ((((x−3)cos θ+(y+6)sin θ)),(((3−x)sin θ+(y+6)cos θ)) ) ⇒2((x−3)cos θ+(y+6)sin θ+3)−3((3−x)sin θ+(y+6)cos θ−6)+a=0 (2x−6)cos θ+(2y+12)sin θ+6−(9−3x)sin θ−(3y+18)cos θ+18+a=0 ⇒2cos θ.x+2sin θ.y+6−6cos θ+12sin θ+3sin θ.x−3cos θ.y−9sin θ−18cos θ+18+a=0 ⇒(2cos θ+3sin θ).x+(2sin θ−3cos θ)y+24+a−24cos θ+3sin θ=0 { ((2cos θ+3sin θ=2...(×3))),((2sin θ−3cos θ=3...(×2))),((24+a−24cos θ+3sin θ=25)) :} { ((6cos θ+9sin θ=6)),((−6cos θ+4sin θ=6)) :} ⇒sin θ=((12)/(13)) & 2cos θ=2−((36)/(13)) cos θ=−(5/(13)) then tan θ = −((12)/5) ⇒a=1+24cos θ−3sin θ ⇒a=1+24(−(5/(13)))−3(((12)/(13))) ⇒a=1−((120)/(13))−((36)/(13)) = ((13−156)/(13))=−((143)/(13))=−11$
$clockwise for θ$ $(\begin{matrix} x^{'} - 3 \\ y^{^{'}} + 6 \end{matrix}) = (\begin{matrix} \cos θ \sin θ \\ - \sin θ \cos θ \end{matrix}) (\begin{matrix} x - 3 \\ y + 6 \end{matrix})$ $(\begin{matrix} x^{'} - 3 \\ y^{'} + 6 \end{matrix}) = (\begin{matrix} (x - 3) \cos θ + (y + 6) \sin θ \\ (3 - x) \sin θ + (y + 6) \cos θ \end{matrix})$ $\Rightarrow 2 ((x - 3) \cos θ + (y + 6) \sin θ + 3) - 3 ((3 - x) \sin θ + (y + 6) \cos θ - 6) + a = 0$ $(2 x - 6) \cos θ + (2 y + 12) \sin θ + 6 - (9 - 3 x) \sin θ - (3 y + 18) \cos θ + 18 + a = 0$ $\Rightarrow 2 \cos θ . x + 2 \sin θ . y + 6 - 6 \cos θ + 12 \sin θ + 3 \sin θ . x - 3 \cos θ . y - 9 \sin θ - 18 \cos θ + 18 + a = 0$ $\Rightarrow (2 \cos θ + 3 \sin θ) . x + (2 \sin θ - 3 \cos θ) y + 24 + a - 24 \cos θ + 3 \sin θ = 0$ ${\begin{cases} 2 \cos θ + 3 \sin θ = 2 . . . (\times 3) \\ 2 \sin θ - 3 \cos θ = 3 . . . (\times 2) \\ 24 + a - 24 \cos θ + 3 \sin θ = 25 \end{cases}$ ${\begin{cases} 6 \cos θ + 9 \sin θ = 6 \\ - 6 \cos θ + 4 \sin θ = 6 \end{cases} \Rightarrow \sin θ = \frac{12}{13} & 2 \cos θ = 2 - \frac{36}{13}$ $\cos θ = - \frac{5}{13} then \tan θ = - \frac{12}{5}$ $\Rightarrow a = 1 + 24 \cos θ - 3 \sin θ$ $\Rightarrow a = 1 + 24 (- \frac{5}{13}) - 3 (\frac{12}{13})$ $\Rightarrow a = 1 - \frac{120}{13} - \frac{36}{13} = \frac{13 - 156}{13} = - \frac{143}{13} = - 11$
Commented by mr W last updated on 04/May/21

$a x + b y + c = 0 \equiv p x + q y + r \Rightarrow a = p, b = q, c = r$ $m e a n s a l s o$ $a x + b y + c = 0 \equiv - p x - q y - r = 0$ $i . e . a = - p, b = - q, c = - r$ $t h e r e f o r e w e h a v e t w o s o l u t i o n s .$
Commented by EDWIN88 last updated on 04/May/21

$yes second solution from$ $distance point (3, - 6) to line first and second line$
Commented by mr W last updated on 05/May/21

Commented by liberty last updated on 05/May/21

$d [(3, - 6), 2 x + 3 y + 25 = 0] = d [(3, - 6), 2 x - 3 y + a = 0]$ $\frac{∣ 6 - 18 + 25 ∣}{\sqrt{13}} = \frac{∣ 6 + 18 + a ∣}{\sqrt{13}}$ $\Rightarrow 13 = ∣ 24 + a ∣ \to {\begin{cases} 24 + a = 13 \to a = - 11 \\ 24 + a = - 13 \to a = - 37 \end{cases}$
	Answered by mr W last updated on 04/May/21

	$f i r s t p o s s i b i l i t y :$ $\tan θ = \tan (π - ϕ) = - \tan ϕ =$ $= - \frac{2 \times \frac{2}{3}}{1 - {(\frac{2}{3})}^{2}} = - \frac{12}{5}$ $2 x + 2 (- 6) + 25 = 2 x - 3 (- 6) + a$ $\Rightarrow a = - 11$
	Commented by mr W last updated on 04/May/21

	Commented by mr W last updated on 04/May/21

	$s e c o n d p o s s i b i l i t y :$ $\tan θ = \tan (2 π - ϕ) = - \tan ϕ = - \frac{12}{5}$ $2 \times 3 + 2 y + 25 = 2 \times 3 - 3 y + a$ $y = - \frac{25 + 2 x}{3} = - \frac{25 + 2 \times 3}{3}$ $y = \frac{a + 2 x}{3} = \frac{a + 2 \times 3}{3}$ $\Rightarrow a = - 37$
	Commented by mr W last updated on 04/May/21

	Answered by mr W last updated on 04/May/21

	$g e n e r a l m e t h o d w i t h o u t u s i n g a n y$ $g e o m e t r y o r g r a p h :$ $x^{'} = (x - 3) \cos θ - (y + 6) \sin θ + 3$ $y^{'} = (x - 3) \sin θ + (y + 6) \cos θ - 6$ $2 [(x - 3) \cos θ - (y + 6) \sin θ + 3] + 3 [(x - 3) \sin θ + (y + 6) \cos θ - 6] + 25 = 0$ $2 x \cos θ - 2 y \sin θ + 3 x \sin θ - 21 \sin θ + 3 y \cos θ + 12 \cos θ + 13 = 0$ $(2 \cos θ + 3 \sin θ) x - (2 \sin θ - 3 \cos θ) y - 21 \sin θ + 12 \cos θ + 13 = 0$ $\equiv 2 x - 3 y + a = 0$ $c a s e 1 :$ $\Rightarrow 2 \cos θ + 3 \sin θ = 2 . . . (i)$ $\Rightarrow 2 \sin θ - 3 \cos θ = 3 . . . (i i i)$ $\Rightarrow - 21 \sin θ + 12 \cos θ + 13 = a . . . (i i i)$ $f r o m (i) a n d (i i) :$ $\sin θ = \frac{12}{13}$ $\cos θ = - \frac{5}{13}$ $s i n c e {(\frac{12}{13})}^{2} + {(- \frac{5}{13})}^{2} = 1 \Rightarrow s o l u t i o n i s$ $c o n s i s t e n t .$ $\Rightarrow \tan θ = - \frac{12}{5} \Rightarrow θ = π - \tan^{- 1} \frac{12}{5}$ $f r o m (i i i) :$ $a = - 21 \times \frac{12}{13} - 12 \times \frac{5}{13} + 13 = - 11$ $c a s e 2 :$ $\Rightarrow 2 \cos θ + 3 \sin θ = - 2 . . . (i)$ $\Rightarrow 2 \sin θ - 3 \cos θ = - 3 . . . (i i i)$ $\Rightarrow - 21 \sin θ + 12 \cos θ + 13 = - a . . . (i i i)$ $f r o m (i) a n d (i i) :$ $\sin θ = - \frac{12}{13}$ $\cos θ = \frac{5}{13}$ $\Rightarrow \tan θ = - \frac{12}{5} \Rightarrow θ = 2 π - \tan^{- 1} \frac{12}{5}$ $f r o m (i i i) :$ $a = - 21 \times \frac{12}{13} - 12 \times \frac{5}{13} - 13 = - 37$
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