Question

The diagram shows the rectangle ABCD, where A is (3,2) and B is (1,6).

Answer 1

(i)
$A(3,2),B(1,6) \\ \\m_{AB}= \frac{6-2}{1-3}=-2 \\ \\m_{BC}m_{AB}=-1 \\ \\m_{BC}= \frac{-1}{m_{AB}} = \frac{-1}{-2}= \frac{1}{2} \\ \\B(1,6)\Rightarrow x_1=1,y_1=6 \\ \\BC:y-y_1=m_{BC}(x-x_1) \\ \\y-6= \frac{1}{2}(x-1) \\ \\2y-12=x-1 \\x-2y+11=0$

(ii)
$AC:y=x-1 \\BC:x-2y+11=0 \\ \\C=AC\cap BC \\ \\x-2(x-1)+11=0 \\x-2x+2+11=0 \\x=13 \\y=x-1=13-1=12 \\C(13,12)$

(iii)
$A(3,2)B(1,6),C(13,12) \\P=2(d(A,B)+d(B,C)) \\ \\d(A,B)= \sqrt{(1-3)^2+(6-2)^2}= \sqrt{20} =2 \sqrt{5} \\ \\d(B,C)= \sqrt{(13-1)^2+(12-6)^2}= \sqrt{180}= 6\sqrt{5} \\ \\P=2(2 \sqrt{5}+6 \sqrt{5})=2\times8 \sqrt{5} =16 \sqrt{5}$