A, B and C are points such that AB is parallel to the line with equation $y + rac{1}{ ext{surd}(3)}x = 0$ and BC makes an angle of 150° with the positive direction of the x-axis - Scottish Highers Maths - Question 9 - 2015

The angle made by line BC with the positive direction of the x-axis is given as 150°. The gradient of a line can be determined using the tangent of the angle:

$m_{BC} = \tan(150°) = \tan(180° - 30°) = -\tan(30°) = -\frac{1}{\sqrt{3}}$

From the calculations, we find that:

• The gradient of AB, $m_{AB} = -\frac{1}{\sqrt{3}}$
• The gradient of BC, $m_{BC} = -\frac{1}{\sqrt{3}}$

Since both gradients are equal ( $m_{AB} = m_{BC}$), it is concluded that the points A, B, and C are collinear.