Hint
Write \(\displaystyle P=(t,0)\) and equate squared distances; this avoids unnecessary square roots.
Worked solution
(a)
Average the corresponding coordinates: \(\displaystyle M= (\frac{- 4+ 8}{2}, \frac{3- 5}{2})= (2, - 1)\).
Answer: \(M=(2,-1)\).
(b)
For \(\displaystyle P=(t,0), PA^{2}=(t+4)^{2}+9\) and \(\displaystyle PB^{2}=(t-8)^{2}+25\). Equality gives \(\displaystyle 8t+25=-16t+89\), so \(\displaystyle 24t=64\) and \(\displaystyle t= \frac{8}{3}\).
Answer: \(\displaystyle P= (\frac{8}{3}, 0)\).
(c)
From M to P the differences are \(\displaystyle \frac{2}{3}\) and 1. Therefore \(\displaystyle MP= \sqrt{\frac{4}{9}+ 1}= \frac{\sqrt{13}}{3}\).
Answer: \(\displaystyle MP= \frac{\sqrt{13}}{3}\).
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 2 marks
- 1 mark: Average the corresponding x- and y-coordinates.
- 1 mark: Give \(\displaystyle M=(2,-1)\).
Alternative methods: A correct midpoint vector calculation is accepted.
Common error: Changing the sign of the negative y-coordinate before averaging.
(b) · 3 marks
- 1 mark: Form the correct squared-distance equation.
- 1 mark: Solve the resulting linear equation to \(\displaystyle t= \frac{8}{3}\).
- 1 mark: Give the paired coordinates \(\displaystyle (\frac{8}{3}, 0)\).
Alternative methods: Intersecting the perpendicular bisector of AB with \(\displaystyle y=0\) is accepted.
Common error: Using the x-coordinate of the midpoint as the axis intersection without checking distances.
(c) · 2 marks
- 1 mark: Apply the distance formula to M and P.
- 1 mark: Give the exact length \(\displaystyle \frac{\sqrt{13}}{3}\); method credit may follow through two earlier coordinate pairs.
Alternative methods: An equivalent exact radical form is accepted.
Common error: Adding the two coordinate differences rather than their squares.
Total: 7 marks.