The equations of six lines are given: Line Equation h x = 3 - y i 2x - 4y = 3 k y = - 1/2(2x - 7) l 4x - 2y - 5 = 0 m x + 3y - 10 = 0 n sqrt{3}x + y - 10 = 0 (a) Complete the table below by matching each description given to one or more of the lines - Leaving Cert Mathematics - Question 3 - 2013

This aligns with line l as well. When substituting x = 0 in its equation, we get y = -2.5.

This is identified as line h, represented by the equation $x + y = 3$, which yields equal x and y intercepts of 3.

This corresponds to line m. The slope can be calculated using the tangent of the angle: $m = an(150°) = -\frac{1}{\sqrt{3}}$, matching the equation of line m.

This can be seen with lines i and k. The slopes of these lines are negative reciprocals of each other.

To find the acute angle between lines m and n, we first determine their slopes:

$m_i = -\frac{1}{\sqrt{3}}$
$m_n = \sqrt{3}$

Using the formula for the tangent of the angle between two lines:

$\tan(\theta) = \frac{m_i - m_n}{1 + m_i m_n}$
Substituting the slopes, we get:

$\tan(\theta) = \frac{-\frac{1}{\sqrt{3}} - \sqrt{3}}{1 + ( -\frac{1}{\sqrt{3}})(\sqrt{3})} = \frac{-\frac{1}{\sqrt{3}} - \sqrt{3}}{1 - 1}$
Thus, we find: $\tan(\theta) = 1/\sqrt{3}$
Therefore, the acute angle $\theta = 30°$.