In the diagram, A(1, 2), B(8, 5), C(13, 7) and D(13, 2) are four points.

1. Find the length of $AB$ using the coordinates of $A$ and $B$.

For $A(1,2)$ and $B(8,5)$,
$AB=\sqrt{(8-1)^2+(5-2)^2}=\sqrt{7^2+3^2}=\sqrt{58}.$

2. Find the length of $BC$, add the two lengths, and round to 3 significant figures.

For $B(8,5)$ and $C(13,7)$,
$BC=\sqrt{(13-8)^2+(7-5)^2}=\sqrt{5^2+2^2}=\sqrt{29}.$
Hence,
$AB+BC=\sqrt{58}+\sqrt{29}=13.0005\ldots=13.0\text{ correct to 3 significant figures}.$

1. Find the direct length $AC$ using the coordinates of $A$ and $C$.

$AC=\sqrt{(13-1)^2+(7-2)^2}$
$=\sqrt{12^2+5^2}$
$=\sqrt{169}=13$

2. Compare $AC$ with the result from part (i) to justify that $A$, $B$ and $C$ lie on a straight line.

From part (i),
$AB+BC=13.0\text{ correct to 3 s.f.}$
Also,
$AC=13=13.0\text{ correct to 3 s.f.}$
Thus,
$AB+BC=AC.$
Therefore $A$, $B$ and $C$ are collinear, with $B$ lying on $AC$.

3. Determine the sum of interior angles of figure $ABCD$.

Since $A$, $B$ and $C$ are collinear, figure $ABCD$ has only three effective sides: $AC$, $CD$ and $DA$.

Hence $ABCD$ is effectively a triangle, so the sum of its interior angles is
$180^\circ.$