(a) P is the point (1,7). Q is the point (5,-5).
(i) Find $\overrightarrow{PQ}$.
(ii) Show that $\left|\overrightarrow{OP}\right| = \left|\overrightarrow{OQ}\right|$.
(iii) PQ is a chord of a circle with centre O. Calculate the circumference of this circle.
(iv) PQ is the diameter of a different circle with centre R. Find the coordinates of R.
(v) Find the equation of the perpendicular bisector of PQ. Give your answer in the form $y = mx + c$.
(b) The position vector of A is a. The position vector of B is b.
M is a point on AB such that AM:MB = 2:3. Find, in terms of a and b, the position vector of M. Give your answer in its simplest form.
▶️ Answer/Explanation
(a)(i) Ans: $\begin{pmatrix} 4 \\ -12 \end{pmatrix}$
$\overrightarrow{PQ} = Q – P = (5-1, -5-7) = (4, -12)$.
(a)(ii) Ans: Both magnitudes are $\sqrt{50}$.
For $P(1,7)$, $|\overrightarrow{OP}| = \sqrt{1^2 + 7^2} = \sqrt{50}$.
For $Q(5,-5)$, $|\overrightarrow{OQ}| = \sqrt{5^2 + (-5)^2} = \sqrt{50}$.
(a)(iii) Ans: $44.4$ or $44.42$ to $44.43$.
Circumference $= 2\pi \times \sqrt{50} \approx 44.42$.
(a)(iv) Ans: $(3, 1)$.
Midpoint $R = \left(\frac{1+5}{2}, \frac{7+(-5)}{2}\right) = (3, 1)$.
(a)(v) Ans: $y = \frac{1}{3}x$.
Slope of PQ is $-3$, so perpendicular slope is $\frac{1}{3}$.
Equation through midpoint $(3,1)$: $y – 1 = \frac{1}{3}(x – 3) \Rightarrow y = \frac{1}{3}x$.
(b) Ans: $\frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}$.
Using section formula, $\overrightarrow{OM} = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}$.
Points \( M(4, 1) \) and \( N(-2, -7) \).
(a) Find the length of \( MN \).
(b) Find the gradient of \( MN \).
(c) Find the equation of the perpendicular bisector of \( MN \).
▶️ Answer/Explanation
(a) 10
Using distance formula: √[(-2-4)² + (-7-1)²] = √(36+64) = √100 = 10.
(b) 4/3
Gradient = (y₂-y₁)/(x₂-x₁) = (-7-1)/(-2-4) = -8/-6 = 4/3.
(c) y=(-3/4)x-9/4
Midpoint is (1,-3). Perpendicular gradient is -3/4 (negative reciprocal). Equation: y+3 = -3/4(x-1).