Solution:

\( m_{AC} \rightarrow \infty \)
\( m_{PD} = 0 \)
\( D\left( \frac{a+a}{2}, \frac{b+3}{2} \right) \)
\( D\left( a, \frac{b+3}{2} \right) \)
\( m_{PD} = 0 \)
\( \frac{b+3}{2} - 1 = 0 \)
\( b+3-2 = 0 \)
\( b = -1 \)
\( E\left( \dfrac{b+a}{2}, \dfrac{5+b}{2} \right) = \left( \dfrac{af}{2}, 2 \right) \)
\( m_{CB}. m_{EP} = -1 \)
\( \left( \dfrac{5-b}{b-a} \right) \left( \dfrac{2-1}{\dfrac{a-1}{2}-1} \right) = -1 \)
\( \left( \frac{6}{-1-a} \right) \left( \frac{2}{a-3} \right) = -1 \)
\( 12 = (1+a)(a-3) \)
\( 12 = a^{2}-3a+a-3 \)
\( \Rightarrow a^{2}-2a-15 = 0 \)
\( (a-5)(a+3)=0 \)
\( a=5 \) or \( a=-3 \)
Given \( ab>0 \)
\( a(-1)>0 \)
\( -a>0 \)
\( a<0 \)
\( a=-3 \) Accept

AP line \( A(-3,3) \, P(1,1) \)
\( y-1=\left(\frac{3-1}{-3-1}\right)(x-1) \)
\( -2y+2=x-1 \)
\( \Rightarrow x+2y=3 \) Appling .....\( (1) \)
Line \( BC \) \( B(-1,5) \)
\( C(-3,-1) \)
\( (y-5)=\frac{6}{2}(x+1) \)
\( y-5=3x+3 \)
\( y=3x+8 \) ....\( (2) \)
Solving \( (1) \) and \( (2) \)
\( x+2(3x+8)=3 \)
\( \Rightarrow 7x+16=3 \)
\( 7x=-13 \)
\( x=-\frac{13}{7} \)
\( y=3\left(-\frac{13}{7}\right)+8 \)
\( =\frac{-39+56}{7} \)
\( y=\frac{17}{7} \)
\( x+y=\frac{-13+17}{7}=\frac{4}{7} \)

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