Answer: \(AC=2\sqrt{34}\); \(y=1\); \(y-1=-\frac35(x-8)\); \(\overrightarrow{EB}=\binom{-5}{3}\), \(\overrightarrow{ED}=\binom{5}{-3}\).

(a) Use the distance formula:

\(AC=\sqrt{(11-5)^2+(6-(-4))^2}.\)

So

\(AC=\sqrt{6^2+10^2}=\sqrt{136}=2\sqrt{34}.\)

(b)(i) The centre of the square is the midpoint of diagonal \(AC\).

Thus

Hence

\(y=1.\)

(b)(ii) The gradient of \(AC\) is

\(\frac{6-(-4)}{11-5}=\frac{10}{6}=\frac53.\)

Since diagonals of a square are perpendicular, the gradient of \(BD\) is

\(-\frac35.\)

The line \(BD\) passes through \(E(8,1)\), so its equation is

\(y-1=-\frac35(x-8).\)

(b)(iii) A vector perpendicular to \(\overrightarrow{AC}=\binom6{10}\) with the same half-diagonal length is

\(\binom{-5}{3}\)

or its negative.

Since the \(x\)-coordinate of \(B\) is less than the \(x\)-coordinate of \(D\),