Step-by-Step Solution
Key Concept: Case study on coordinate geometry.
(a) Find the coordinates of the checkpoint (the midpoint of AB). [1 Mark]
Midpoint $=\left(\dfrac{5+13}{2},\dfrac{2+10}{2}\right)=(9,6)$. [1.0 Mark]
(b) Find the total distance the character must travel from A to the exit B. [1 Mark]
$AB=\sqrt{(13-5)^2+(10-2)^2}=\sqrt{64+64}=\sqrt{128}=8\sqrt2$ units. [1.0 Mark]
(c) Confirm that the distance from A to the checkpoint equals the distance from the checkpoint to B. [1 Mark]
Distance from $A(5,2)$ to checkpoint $(9,6)$: $\sqrt{16+16}=4\sqrt2$. Distance from checkpoint to $B(13,10)$: $\sqrt{16+16}=4\sqrt2$. Both are equal, confirming the checkpoint is indeed the midpoint. [1.0 Mark]
(d) If a bonus item is placed at the point dividing AB in the ratio 3:1 (from A), find its coordinates. [1 Mark]
$x=\dfrac{3(13)+1(5)}{3+1}=\dfrac{39+5}{4}=11$; $y=\dfrac{3(10)+1(2)}{4}=\dfrac{30+2}{4}=8$. So the bonus item is at $(11,8)$. [1.0 Mark]
Correct Answer: Midpoint $=\left(\dfrac{5+13}{2},\dfrac{2+10}{2}\right)=(9,6)$. [1.0 Mark] | $AB=\sqrt{(13-5)^2+(10-2)^2}=\sqrt{64+64}=\sqrt{128}=8\sqrt2$ units. [1.0 Mark] | Distance from $A(5,2)$ to checkpoint $(9,6)$: $\sqrt{16+16}=4\sqrt2$. Distance from checkpoint to $B(13,10)$: $\sqrt{16+16}=4\sqrt2$. Both are equal, confirming the checkpoint is indeed the midpoint. [1.0 Mark] | $x=\dfrac{3(13)+1(5)}{3+1}=\dfrac{39+5}{4}=11$; $y=\dfrac{3(10)+1(2)}{4}=\dfrac{30+2}{4}=8$. So the bonus item is at $(11,8)$. [1.0 Mark]