Problems in Plane Analytic Geometry: Difficult Problems with Solutions

Solution:
[tex]D:2x-y=1, D':(a-1)x+2y=x-2[/tex]
[tex]D\parallel D' \Rightarrow m_D=m_D'[/tex]
[tex]2x-y=1\rightarrow y=2x-1\rightarrow m_D=2[/tex]
[tex](a-1)x+2y=x-2\rightarrow 2y=x(1-a+1)-2 \rightarrow 2y=x(2-a)-2 \rightarrow[/tex]
[tex]y=x\times \frac{(2-a)}{2}-1\rightarrow m_D'= \frac{(2-a)}{2}[/tex]
[tex]m_D=m_D' \rightarrow \frac{(2-a)}{2}=2 \rightarrow 2-a=4 \rightarrow a=-2[/tex]

Solution:
[tex]D:(2a-3)x+ay-2=2y+3a, D':(3a-1)y+2(3a+1)x=a-1[/tex]
[tex]D\perp D' \Rightarrow m_D\times m_D'=-1[/tex]
[tex](2a-3)x+ay-2=2y+3a \rightarrow 2y-ay=(2a-3)x-2-3a \rightarrow[/tex]
[tex]y(2-a)=(2a-3)x-2-3a \rightarrow y=\frac{(2a-3)}{(2-a)}x-\frac{(2+3a)}{(2-a)}\rightarrow m_D=\frac{(2a-3)}{(2-a)}[/tex]
[tex](3a-1)y+2(3a+1)x=a-1 \rightarrow (1-3a)y=2(3a+1)x+1-a \rightarrow[/tex]
[tex]y=\frac{2(3a+1)}{(1-3a)}x+\frac{1-a}{1-3a} \rightarrow m_D'= \frac{2(3a+1)}{(1-3a)}[/tex]
[tex]\frac{(2a-3)}{(2-a)}\times \frac{2(3a+1)}{(1-3a)}=-1 \rightarrow (2a-3)\times 2(3a+1)=-(2-a)(1-3a) \rightarrow[/tex]
[tex]2(6a^2-7a-3)=-(3a^2-7a+2)\rightarrow 12a^2-14a-6=-3a^2+7a-2 \rightarrow 15a^2-21a-4=0 \rightarrow a_1,a_2=\frac{21\pm\sqrt{21^2+16\times 15}}{30}=\frac{21\pm\sqrt{861}}{30}[/tex]