Hint

A vertical segment has a horizontal perpendicular bisector. The bisector must pass through the midpoint.

Worked solution

(a)

Both endpoints have \(\displaystyle x=3\), so AB is the vertical line \(\displaystyle x=3\). Its midpoint is \(\displaystyle (3, \frac{- 2+ 6}{2})= (3, 2)\).

Answer: \(x=3\); midpoint \((3,2)\).

(b)

It is horizontal because AB is vertical, and it passes through the midpoint (3,2). Hence \(\displaystyle y=2\).

Answer: \(y=2\).

(c)

Set \(\displaystyle 2=2x+6\) to obtain \(\displaystyle x=-2\), so \(\displaystyle Q=(-2,2)\). The differences from A are −5 and 4, giving \(\displaystyle QA=\sqrt{25+16}=\sqrt{41}\). The same value is obtained for QB.

Answer: \(Q=(-2,2)\); \(QA=\sqrt{41}\).

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: State \(\displaystyle x=3\).
• 1 mark: Give midpoint (3,2).

Alternative methods: A clear vertical-line observation is accepted.

Common error: Reporting gradient zero for a vertical line.

(b) · 2 marks

• 1 mark: Identify the required horizontal direction.
• 1 mark: Use the midpoint to give \(\displaystyle y=2\).

Alternative methods: An equal-squared-distance derivation is accepted.

Common error: Giving a perpendicular line through A rather than through the midpoint.

(c) · 3 marks

• 1 mark: Intersect the two lines to obtain \(\displaystyle Q=(-2,2)\).
• 1 mark: Use the correct squared differences from Q to A.
• 1 mark: Give \(\displaystyle QA=\sqrt{41}\).

Alternative methods: Using QB after explaining equidistance is accepted.

Common error: Confusing the midpoint of AB with Q.

Total: 7 marks.