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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

eqn. 2x−3y+a=0 fixes only the inclination of the line. the vertical position will be determined by the value of a. the distance from point (3,−6) must to this line must be the same as to 2x+3y+25=0, so we get two parallel lines with different a.

$${eqn}.\:\mathrm{2}{x}−\mathrm{3}{y}+{a}=\mathrm{0}\:{fixes}\:{only}\:{the} \\ $$$${inclination}\:{of}\:{the}\:{line}.\:{the}\:{vertical} \\ $$$${position}\:{will}\:{be}\:{determined}\:{by}\:{the} \\ $$$${value}\:{of}\:{a}.\:{the}\:{distance}\:{from}\:{point} \\ $$$$\left(\mathrm{3},−\mathrm{6}\right)\:{must}\:{to}\:{this}\:{line}\:{must}\:{be}\:{the} \\ $$$${same}\:{as}\:{to}\:\mathrm{2}{x}+\mathrm{3}{y}+\mathrm{25}=\mathrm{0},\:{so}\:{we}\:{get} \\ $$$${two}\:{parallel}\:{lines}\:{with}\:{different}\:{a}. \\ $$

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

$$\mathrm{clockwise}\:\mathrm{for}\:\theta\: \\ $$$$\begin{pmatrix}{\mathrm{x}'−\mathrm{3}}\\{\mathrm{y}^{'} +\mathrm{6}}\end{pmatrix}\:=\:\begin{pmatrix}{\:\:\:\mathrm{cos}\:\theta\:\:\:\:\mathrm{sin}\:\theta}\\{−\mathrm{sin}\:\theta\:\:\:\mathrm{cos}\:\theta}\end{pmatrix}\:\begin{pmatrix}{\mathrm{x}−\mathrm{3}}\\{\mathrm{y}+\mathrm{6}}\end{pmatrix} \\ $$$$\begin{pmatrix}{\mathrm{x}'−\mathrm{3}}\\{\mathrm{y}'+\mathrm{6}}\end{pmatrix}\:=\:\begin{pmatrix}{\left(\mathrm{x}−\mathrm{3}\right)\mathrm{cos}\:\theta+\left(\mathrm{y}+\mathrm{6}\right)\mathrm{sin}\:\theta}\\{\left(\mathrm{3}−\mathrm{x}\right)\mathrm{sin}\:\theta+\left(\mathrm{y}+\mathrm{6}\right)\mathrm{cos}\:\theta}\end{pmatrix} \\ $$$$\Rightarrow\mathrm{2}\left(\left(\mathrm{x}−\mathrm{3}\right)\mathrm{cos}\:\theta+\left(\mathrm{y}+\mathrm{6}\right)\mathrm{sin}\:\theta+\mathrm{3}\right)−\mathrm{3}\left(\left(\mathrm{3}−\mathrm{x}\right)\mathrm{sin}\:\theta+\left(\mathrm{y}+\mathrm{6}\right)\mathrm{cos}\:\theta−\mathrm{6}\right)+\mathrm{a}=\mathrm{0} \\ $$$$\left(\mathrm{2x}−\mathrm{6}\right)\mathrm{cos}\:\theta+\left(\mathrm{2y}+\mathrm{12}\right)\mathrm{sin}\:\theta+\mathrm{6}−\left(\mathrm{9}−\mathrm{3x}\right)\mathrm{sin}\:\theta−\left(\mathrm{3y}+\mathrm{18}\right)\mathrm{cos}\:\theta+\mathrm{18}+\mathrm{a}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2cos}\:\theta.\mathrm{x}+\mathrm{2sin}\:\theta.\mathrm{y}+\mathrm{6}−\mathrm{6cos}\:\theta+\mathrm{12sin}\:\theta+\mathrm{3sin}\:\theta.\mathrm{x}−\mathrm{3cos}\:\theta.\mathrm{y}−\mathrm{9sin}\:\theta−\mathrm{18cos}\:\theta+\mathrm{18}+\mathrm{a}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2cos}\:\theta+\mathrm{3sin}\:\theta\right).\mathrm{x}+\left(\mathrm{2sin}\:\theta−\mathrm{3cos}\:\theta\right)\mathrm{y}+\mathrm{24}+\mathrm{a}−\mathrm{24cos}\:\theta+\mathrm{3sin}\:\theta=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{2cos}\:\theta+\mathrm{3sin}\:\theta=\mathrm{2}...\left(×\mathrm{3}\right)}\\{\mathrm{2sin}\:\theta−\mathrm{3cos}\:\theta=\mathrm{3}...\left(×\mathrm{2}\right)}\\{\mathrm{24}+\mathrm{a}−\mathrm{24cos}\:\theta+\mathrm{3sin}\:\theta=\mathrm{25}}\end{cases} \\ $$$$\begin{cases}{\mathrm{6cos}\:\theta+\mathrm{9sin}\:\theta=\mathrm{6}}\\{−\mathrm{6cos}\:\theta+\mathrm{4sin}\:\theta=\mathrm{6}}\end{cases}\:\Rightarrow\mathrm{sin}\:\theta=\frac{\mathrm{12}}{\mathrm{13}}\:\&\:\mathrm{2cos}\:\theta=\mathrm{2}−\frac{\mathrm{36}}{\mathrm{13}} \\ $$$$\mathrm{cos}\:\theta=−\frac{\mathrm{5}}{\mathrm{13}}\:\mathrm{then}\:\mathrm{tan}\:\theta\:=\:−\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{a}=\mathrm{1}+\mathrm{24cos}\:\theta−\mathrm{3sin}\:\theta \\ $$$$\Rightarrow\mathrm{a}=\mathrm{1}+\mathrm{24}\left(−\frac{\mathrm{5}}{\mathrm{13}}\right)−\mathrm{3}\left(\frac{\mathrm{12}}{\mathrm{13}}\right) \\ $$$$\Rightarrow\mathrm{a}=\mathrm{1}−\frac{\mathrm{120}}{\mathrm{13}}−\frac{\mathrm{36}}{\mathrm{13}}\:=\:\frac{\mathrm{13}−\mathrm{156}}{\mathrm{13}}=−\frac{\mathrm{143}}{\mathrm{13}}=−\mathrm{11} \\ $$

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

ax+by+c=0≡px+qy+r ⇒a=p,b=q,c=r means also ax+by+c=0≡−px−qy−r=0 i.e. a=−p,b=−q,c=−r therefore we have two solutions.

$${ax}+{by}+{c}=\mathrm{0}\equiv{px}+{qy}+{r}\:\Rightarrow{a}={p},{b}={q},{c}={r} \\ $$$${means}\:{also} \\ $$$${ax}+{by}+{c}=\mathrm{0}\equiv−{px}−{qy}−{r}=\mathrm{0} \\ $$$${i}.{e}.\:{a}=−{p},{b}=−{q},{c}=−{r} \\ $$$${therefore}\:{we}\:{have}\:{two}\:{solutions}. \\ $$

Commented by EDWIN88 last updated on 04/May/21

yes second solution from distance point(3,−6) to line first and second line

$$\mathrm{yes}\:\mathrm{second}\:\mathrm{solution}\:\mathrm{from}\: \\ $$$$\:\mathrm{distance}\:\mathrm{point}\left(\mathrm{3},−\mathrm{6}\right)\:\mathrm{to}\:\mathrm{line}\:\mathrm{first}\:\mathrm{and}\:\mathrm{second}\:\mathrm{line} \\ $$

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

Commented by liberty last updated on 05/May/21

d [(3,−6),2x+3y+25=0 ]= d[(3,−6),2x−3y+a=0 ] ((∣6−18+25∣)/( (√(13)))) = ((∣6+18+a∣)/( (√(13)))) ⇒13 = ∣24+a∣ → { ((24+a=13→a=−11)),((24+a=−13→a=−37)) :}

$$\:\mathrm{d}\:\left[\left(\mathrm{3},−\mathrm{6}\right),\mathrm{2x}+\mathrm{3y}+\mathrm{25}=\mathrm{0}\:\right]=\:\mathrm{d}\left[\left(\mathrm{3},−\mathrm{6}\right),\mathrm{2x}−\mathrm{3y}+\mathrm{a}=\mathrm{0}\:\right] \\ $$$$\:\frac{\mid\mathrm{6}−\mathrm{18}+\mathrm{25}\mid}{\:\sqrt{\mathrm{13}}}\:=\:\frac{\mid\mathrm{6}+\mathrm{18}+\mathrm{a}\mid}{\:\sqrt{\mathrm{13}}} \\ $$$$\Rightarrow\mathrm{13}\:=\:\mid\mathrm{24}+\mathrm{a}\mid\:\rightarrow\begin{cases}{\mathrm{24}+\mathrm{a}=\mathrm{13}\rightarrow\mathrm{a}=−\mathrm{11}}\\{\mathrm{24}+\mathrm{a}=−\mathrm{13}\rightarrow\mathrm{a}=−\mathrm{37}}\end{cases} \\ $$

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

first possibility: tan θ=tan (π−φ)=−tan φ= =−((2×(2/3))/(1−((2/3))^2 ))=−((12)/5) 2x+2(−6)+25=2x−3(−6)+a ⇒a=−11

$${first}\:{possibility}: \\ $$$$\mathrm{tan}\:\theta=\mathrm{tan}\:\left(\pi−\phi\right)=−\mathrm{tan}\:\phi= \\ $$$$=−\frac{\mathrm{2}×\frac{\mathrm{2}}{\mathrm{3}}}{\mathrm{1}−\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} }=−\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\mathrm{2}{x}+\mathrm{2}\left(−\mathrm{6}\right)+\mathrm{25}=\mathrm{2}{x}−\mathrm{3}\left(−\mathrm{6}\right)+{a} \\ $$$$\Rightarrow{a}=−\mathrm{11} \\ $$

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

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

second possibility: tan θ=tan (2π−φ)=−tan φ=−((12)/5) 2×3+2y+25=2×3−3y+a y=−((25+2x)/3)=−((25+2×3)/3) y=((a+2x)/3)=((a+2×3)/3) ⇒a=−37

$${second}\:{possibility}: \\ $$$$\mathrm{tan}\:\theta=\mathrm{tan}\:\left(\mathrm{2}\pi−\phi\right)=−\mathrm{tan}\:\phi=−\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\mathrm{2}×\mathrm{3}+\mathrm{2}{y}+\mathrm{25}=\mathrm{2}×\mathrm{3}−\mathrm{3}{y}+{a} \\ $$$${y}=−\frac{\mathrm{25}+\mathrm{2}{x}}{\mathrm{3}}=−\frac{\mathrm{25}+\mathrm{2}×\mathrm{3}}{\mathrm{3}} \\ $$$${y}=\frac{{a}+\mathrm{2}{x}}{\mathrm{3}}=\frac{{a}+\mathrm{2}×\mathrm{3}}{\mathrm{3}} \\ $$$$\Rightarrow{a}=−\mathrm{37} \\ $$

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

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

general method without using any geometry or graph: 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 2xcos θ−2ysin θ+3xsin θ−21sin θ+3ycos θ+12cos θ+13=0 (2cos θ+3sin θ)x−(2sin θ−3cos θ)y−21sin θ+12cos θ+13=0 ≡2x−3y+a=0 case 1: ⇒2cos θ+3sin θ=2 ...(i) ⇒2sin θ−3cos θ=3 ...(iii) ⇒−21sin θ+12cos θ+13=a ...(iii) from (i) and (ii): sin θ=((12)/(13)) cos θ=−(5/(13)) since (((12)/(13)))^2 +(−(5/(13)))^2 =1 ⇒solution is consistent. ⇒tan θ=−((12)/5) ⇒θ=π−tan^(−1) ((12)/5) from (iii): a=−21×((12)/(13))−12×(5/(13))+13=−11 case 2: ⇒2cos θ+3sin θ=−2 ...(i) ⇒2sin θ−3cos θ=−3 ...(iii) ⇒−21sin θ+12cos θ+13=−a ...(iii) from (i) and (ii): sin θ=−((12)/(13)) cos θ=(5/(13)) ⇒tan θ=−((12)/5) ⇒θ=2π−tan^(−1) ((12)/5) from (iii): a=−21×((12)/(13))−12×(5/(13))−13=−37

$${general}\:{method}\:{without}\:{using}\:{any} \\ $$$${geometry}\:{or}\:{graph}: \\ $$$$ \\ $$$${x}'=\left({x}−\mathrm{3}\right)\mathrm{cos}\:\theta−\left({y}+\mathrm{6}\right)\mathrm{sin}\:\theta+\mathrm{3} \\ $$$${y}'=\left({x}−\mathrm{3}\right)\mathrm{sin}\:\theta+\left({y}+\mathrm{6}\right)\mathrm{cos}\:\theta−\mathrm{6} \\ $$$$ \\ $$$$\mathrm{2}\left[\left({x}−\mathrm{3}\right)\mathrm{cos}\:\theta−\left({y}+\mathrm{6}\right)\mathrm{sin}\:\theta+\mathrm{3}\right]+\mathrm{3}\left[\left({x}−\mathrm{3}\right)\mathrm{sin}\:\theta+\left({y}+\mathrm{6}\right)\mathrm{cos}\:\theta−\mathrm{6}\right]+\mathrm{25}=\mathrm{0} \\ $$$$\mathrm{2}{x}\mathrm{cos}\:\theta−\mathrm{2}{y}\mathrm{sin}\:\theta+\mathrm{3}{x}\mathrm{sin}\:\theta−\mathrm{21sin}\:\theta+\mathrm{3}{y}\mathrm{cos}\:\theta+\mathrm{12cos}\:\theta+\mathrm{13}=\mathrm{0} \\ $$$$ \\ $$$$\left(\mathrm{2cos}\:\theta+\mathrm{3sin}\:\theta\right){x}−\left(\mathrm{2sin}\:\theta−\mathrm{3cos}\:\theta\right){y}−\mathrm{21sin}\:\theta+\mathrm{12cos}\:\theta+\mathrm{13}=\mathrm{0} \\ $$$$\equiv\mathrm{2}{x}−\mathrm{3}{y}+{a}=\mathrm{0} \\ $$$${case}\:\mathrm{1}: \\ $$$$\Rightarrow\mathrm{2cos}\:\theta+\mathrm{3sin}\:\theta=\mathrm{2}\:\:\:...\left({i}\right) \\ $$$$\Rightarrow\mathrm{2sin}\:\theta−\mathrm{3cos}\:\theta=\mathrm{3}\:\:\:...\left({iii}\right) \\ $$$$\Rightarrow−\mathrm{21sin}\:\theta+\mathrm{12cos}\:\theta+\mathrm{13}={a}\:\:\:...\left({iii}\right) \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{12}}{\mathrm{13}} \\ $$$$\mathrm{cos}\:\theta=−\frac{\mathrm{5}}{\mathrm{13}} \\ $$$${since}\:\left(\frac{\mathrm{12}}{\mathrm{13}}\right)^{\mathrm{2}} +\left(−\frac{\mathrm{5}}{\mathrm{13}}\right)^{\mathrm{2}} =\mathrm{1}\:\Rightarrow{solution}\:{is} \\ $$$${consistent}. \\ $$$$\Rightarrow\mathrm{tan}\:\theta=−\frac{\mathrm{12}}{\mathrm{5}}\:\Rightarrow\theta=\pi−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{12}}{\mathrm{5}} \\ $$$${from}\:\left({iii}\right): \\ $$$${a}=−\mathrm{21}×\frac{\mathrm{12}}{\mathrm{13}}−\mathrm{12}×\frac{\mathrm{5}}{\mathrm{13}}+\mathrm{13}=−\mathrm{11} \\ $$$$ \\ $$$${case}\:\mathrm{2}: \\ $$$$\Rightarrow\mathrm{2cos}\:\theta+\mathrm{3sin}\:\theta=−\mathrm{2}\:\:\:...\left({i}\right) \\ $$$$\Rightarrow\mathrm{2sin}\:\theta−\mathrm{3cos}\:\theta=−\mathrm{3}\:\:\:...\left({iii}\right) \\ $$$$\Rightarrow−\mathrm{21sin}\:\theta+\mathrm{12cos}\:\theta+\mathrm{13}=−{a}\:\:\:...\left({iii}\right) \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$$\mathrm{sin}\:\theta=−\frac{\mathrm{12}}{\mathrm{13}} \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{5}}{\mathrm{13}} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=−\frac{\mathrm{12}}{\mathrm{5}}\:\Rightarrow\theta=\mathrm{2}\pi−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{12}}{\mathrm{5}} \\ $$$${from}\:\left({iii}\right): \\ $$$${a}=−\mathrm{21}×\frac{\mathrm{12}}{\mathrm{13}}−\mathrm{12}×\frac{\mathrm{5}}{\mathrm{13}}−\mathrm{13}=−\mathrm{37} \\ $$

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