(i)
[tex]A(3,2),B(1,6)
\\
\\m_{AB}= \frac{6-2}{1-3}=-2
\\
\\m_{BC}m_{AB}=-1
\\
\\m_{BC}= \frac{-1}{m_{AB}} = \frac{-1}{-2}= \frac{1}{2}
\\
\\B(1,6)\Rightarrow x_1=1,y_1=6
\\
\\BC:y-y_1=m_{BC}(x-x_1)
\\
\\y-6= \frac{1}{2}(x-1)
\\
\\2y-12=x-1
\\x-2y+11=0 [/tex]
(ii)
[tex]AC:y=x-1
\\BC:x-2y+11=0
\\
\\C=AC\cap BC
\\
\\x-2(x-1)+11=0
\\x-2x+2+11=0
\\x=13
\\y=x-1=13-1=12
\\C(13,12)[/tex]
(iii)
[tex]A(3,2)B(1,6),C(13,12)
\\P=2(d(A,B)+d(B,C))
\\
\\d(A,B)= \sqrt{(1-3)^2+(6-2)^2}= \sqrt{20} =2 \sqrt{5}
\\
\\d(B,C)= \sqrt{(13-1)^2+(12-6)^2}= \sqrt{180}= 6\sqrt{5}
\\
\\P=2(2 \sqrt{5}+6 \sqrt{5})=2\times8 \sqrt{5} =16 \sqrt{5}[/tex]
