Find the acute angle between the line [tex]\( L_1 \)[/tex] which passes through the points [tex]\( A(2,2) \)[/tex] and [tex]\( B(8,6) \)[/tex] and the line [tex]\( L_2 \)[/tex] which passes through the points [tex]\( C(0,3) \)[/tex] and [tex]\( D(6,-3) \)[/tex].



Answer :

To find the acute angle between the two lines [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex], we need to follow a series of steps:

1. Calculate the slopes of the lines:

The formula for the slope [tex]\( m \)[/tex] of a line passing through two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]

- For line [tex]\( L_1 \)[/tex] passing through points [tex]\( A(2,2) \)[/tex] and [tex]\( B(8,6) \)[/tex]:
[tex]\[ m_1 = \frac{6 - 2}{8 - 2} = \frac{4}{6} = \frac{2}{3} \approx 0.6667 \][/tex]

- For line [tex]\( L_2 \)[/tex] passing through points [tex]\( C(0,3) \)[/tex] and [tex]\( D(6,-3) \)[/tex]:
[tex]\[ m_2 = \frac{-3 - 3}{6 - 0} = \frac{-6}{6} = -1 \][/tex]

2. Calculate the angle between the two lines:

The angle [tex]\(\theta\)[/tex] between two lines with slopes [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] can be found using the formula:
[tex]\[ \theta = \tan^{-1} \left( \left| \frac{m_1 - m_2}{1 + m_1 \cdot m_2} \right| \right) \][/tex]

Substituting the slopes [tex]\( m_1 \approx 0.6667 \)[/tex] and [tex]\( m_2 = -1 \)[/tex]:

[tex]\[ \theta = \tan^{-1} \left( \left| \frac{0.6667 - (-1)}{1 + 0.6667 \cdot (-1)} \right| \right) = \tan^{-1} \left( \left| \frac{0.6667 + 1}{1 - 0.6667} \right| \right) = \tan^{-1} \left( \left| \frac{1.6667}{0.3333} \right| \right) \approx \tan^{-1}(5) \][/tex]

3. Convert the angle from radians to degrees:

Using a calculator, we can find that:
[tex]\[ \theta \approx 1.3734 \text{ radians} \][/tex]

To convert radians to degrees, we use the conversion factor [tex]\( \frac{180}{\pi} \)[/tex]:
[tex]\[ \theta_{\text{degrees}} = 1.3734 \times \frac{180}{\pi} \approx 78.69^\circ \][/tex]

Therefore, the acute angle between the lines [tex]\( L_1 \)[/tex] and [tex]\( L_2 \)[/tex] is approximately [tex]\( 78.69^\circ \)[/tex].