The coordinates of three points are A(2,−6), B(6,−12), and C(−4, 3) - Leaving Cert Mathematics - Question 1 - 2020

The formula for the perpendicular distance (d) from a point (x_0, y_0) to a line Ax + By + C = 0 is given by: $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

For the line equation: (-\frac{3}{2}x + y + 3 = 0), this gives:

• A = -\frac{3}{2}, B = 1, and C = 3. For point A(2, -6):

$d = \frac{|(-\frac{3}{2})(2) + (1)(-6) + 3|}{\sqrt{(-\frac{3}{2})^2 + 1^2}}$

Calculating: $d = \frac{|-3 + 6 + 3|}{\sqrt{\frac{9}{4} + 1}} = \frac{6}{\sqrt{\frac{13}{4}}} = \frac{6 * 2}{\sqrt{13}} = \frac{12}{\sqrt{13}}$

Since the distance from point A to line BC is valid and calculated, we may conclude that points A, B, and C are collinear, as the area of triangle ABC can be seen to be zero based on the determined perpendicular distance.

For line a, the equation is given as x - 2y + 1 = 0. Rearranging gives: $y = \frac{1}{2}x + \frac{1}{2}$

The slope of line a is thus (m_a = \frac{1}{2}). Line b makes an angle of 60° with the x-axis, thus the slope can be calculated from: $m_b = \tan(60°) = \sqrt{3}$

Using the formula for the angle between two lines with slopes m1 and m2: $\tan(\theta) = \frac{m_2 - m_1}{1 + m_1m_2}$ Let m1 = \frac{1}{2} and m2 = \sqrt{3}:

$\tan(\theta) = \frac{\sqrt{3} - \frac{1}{2}}{1 + (\frac{1}{2})(\sqrt{3})}$ After substituting and simplifying, we can use the arctan to find θ. This can be calculated as follows: $\theta = \tan^{-1}\left(\frac{\frac{\sqrt{3} - \frac{1}{2}}{1 + \frac{1}{2}\sqrt{3}}}{1}\right)$ Evaluating results in: $\theta = 33.435°$