Find the equation of line \( AB \) using points \( A(2, 5) \) and \( B(8, 2) \).
Gradient: \( m = \frac{2 – 5}{8 – 2} = \frac{-3}{6} = -\frac{1}{2} \).
Use point-slope form at \( A(2, 5) \): \( y – 5 = -\frac{1}{2} \times (x – 2) \).
Simplify: \( y – 5 = -\frac{1}{2} \times x + 1 \implies y = -\frac{1}{2} \times x + 6 \).
\( y \)-intercept: 6 (or \( (0, 6) \)) (A1 N1).

[1 mark]

Use gradient formula: \( m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{2 – 5}{8 – 2} = \frac{-3}{6} = -\frac{1}{2} \) (M1 A1 N2).

[2 marks]

On the diagram, line \( AB \) passes through \( A(2, 5) \) and \( B(8, 2) \), sloping downward. At point \( B(8, 2) \), draw a horizontal line along the positive \( x \)-axis (to the right). Label the acute angle between line \( AB \) and the horizontal line as \( \theta \), ensuring it is less than 90°. (A1 N1).

[Image Reference: The diagram shows \( \theta \) as the acute angle at \( B(8, 2) \) between line \( AB \) and the positive \( x \)-axis.]

[1 mark]

Method 1: Tangent Function
Gradient of \( AB \): \( -\frac{1}{2} \). For acute angle \( \theta \), use \( \tan \theta = |\text{gradient}| = \left| -\frac{1}{2} \right| = \frac{1}{2} \) (M1).
\( \theta = \arctan\left(\frac{1}{2}\right) \approx 0.463648 \text{ radians} \approx 26.565^\circ \).
Correct to one decimal place: \( \theta \approx 26.6^\circ \) (A1 N2).

Method 2: Vector Dot Product
Vector along \( AB \): \( \vec{AB} = (8 – 2, 2 – 5) = (6, -3) \). Vector along positive \( x \)-axis: \( (1, 0) \).
Cosine of angle: \( \cos \theta = \frac{\vec{AB} \cdot (1, 0)}{|\vec{AB}| \times |(1, 0)|} = \frac{6 \times 1 + (-3) \times 0}{\sqrt{6^2 + (-3)^2} \times 1} = \frac{6}{\sqrt{45}} = \frac{6}{3 \sqrt{5}} = \frac{2}{\sqrt{5}} \).
Acute angle: \( \theta = \arccos\left(\frac{2}{\sqrt{5}}\right) \approx 26.565^\circ \approx 26.6^\circ \) (M1 A1 N2).

[2 marks]