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(A) If points $A(-5, y)$, $B(2, -2)$, $C(8, 4)$ and $D(x, 5)$ taken in order, form a parallelogram ABCD, then find the values of $x$ and $y$. Hence, find lengths of sides of the parallelogram.
OR
(B) $A(6, -3)$, $B(0, 5)$ and $C(-2, 1)$ are vertices of $\Delta ABC$. Points $P(3, 1)$ and $Q(2, -1)$ lie on sides AB and AC respectively. Check whether $\frac{AP}{PB} = \frac{AQ}{QC}$.
OR
(B) $A(6, -3)$, $B(0, 5)$ and $C(-2, 1)$ are vertices of $\Delta ABC$. Points $P(3, 1)$ and $Q(2, -1)$ lie on sides AB and AC respectively. Check whether $\frac{AP}{PB} = \frac{AQ}{QC}$.
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(A) ABCD is a parallelogram
$\therefore$ Coordinates of mid pt. of BD = Coordinates of mid pt. of AC
$(\frac{2+x}{2}, \frac{-2+5}{2}) = (\frac{8-5}{2}, \frac{4+y}{2})$ [$1$ mark]
Getting $x = 1$ and $y = -1$ [$\frac{1}{2} + \frac{1}{2}$ mark]
$AB = \sqrt{7^2 + (-1)^2} = \sqrt{50}$ or $5\sqrt{2}$ [$\frac{1}{2}$ mark]
$BC = \sqrt{6^2 + 6^2} = \sqrt{72}$ or $6\sqrt{2}$ [$\frac{1}{2}$ mark]
OR
(B) $AP = \sqrt{3^2 + (-4)^2} = 5$ [$\frac{1}{2}$ mark]
$PB = \sqrt{3^2 + (-4)^2} = 5$ [$\frac{1}{2}$ mark]
$AQ = \sqrt{4^2 + (-2)^2} = 2\sqrt{5}$ [$\frac{1}{2}$ mark]
$QC = \sqrt{4^2 + (-2)^2} = 2\sqrt{5}$ [$\frac{1}{2}$ mark]
So $\frac{AP}{PB} = 1$ and $\frac{AQ}{QC} = 1$
Therefore $\frac{AP}{PB} = \frac{AQ}{QC}$ [$1$ mark]
$\therefore$ Coordinates of mid pt. of BD = Coordinates of mid pt. of AC
$(\frac{2+x}{2}, \frac{-2+5}{2}) = (\frac{8-5}{2}, \frac{4+y}{2})$ [$1$ mark]
Getting $x = 1$ and $y = -1$ [$\frac{1}{2} + \frac{1}{2}$ mark]
$AB = \sqrt{7^2 + (-1)^2} = \sqrt{50}$ or $5\sqrt{2}$ [$\frac{1}{2}$ mark]
$BC = \sqrt{6^2 + 6^2} = \sqrt{72}$ or $6\sqrt{2}$ [$\frac{1}{2}$ mark]
OR
(B) $AP = \sqrt{3^2 + (-4)^2} = 5$ [$\frac{1}{2}$ mark]
$PB = \sqrt{3^2 + (-4)^2} = 5$ [$\frac{1}{2}$ mark]
$AQ = \sqrt{4^2 + (-2)^2} = 2\sqrt{5}$ [$\frac{1}{2}$ mark]
$QC = \sqrt{4^2 + (-2)^2} = 2\sqrt{5}$ [$\frac{1}{2}$ mark]
So $\frac{AP}{PB} = 1$ and $\frac{AQ}{QC} = 1$
Therefore $\frac{AP}{PB} = \frac{AQ}{QC}$ [$1$ mark]