4.4 Q-10

Question Statement

We are asked to find the angle between two lines, $\ell_1$ and $\ell_2$, in multiple cases. Specifically, the lines are defined by the following points:

(a)

• Line $\ell_1$ joins points $(2,7)$ and $(7,10)$
• Line $\ell_2$ joins points $(1,1)$ and $(-5,3)$

Also, for each case, we are to find the acute angle between the lines.

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Background and Explanation

To find the angle between two lines, we need to use the slopes of the lines. The slope of a line through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

The formula for the angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is:

$$\tan \theta = \frac{|m_1 - m_2|}{1 + m_1 m_2}$$

Once we compute $\tan \theta$, we can find $\theta$ using the inverse tangent function.

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Solution

(a) Line 1 and Line 2:

We first calculate the slopes of $\ell_1$ and $\ell_2$:

1. Slope of $\ell_1$:
   • Points: $(2,7)$ and $(7,10)$
   • Using the slope formula:

$$m_1 = \frac{10 - 7}{7 - 2} = \frac{3}{5}$$

2. Slope of $\ell_2$:
   • Points: $(1,1)$ and $(-5,3)$
   • Using the slope formula:

$$m_2 = \frac{3 - 1}{-5 - 1} = \frac{2}{-6} = -\frac{1}{3}$$

Now, we can find the angle between the lines:

$$\tan \theta = \frac{|m_1 - m_2|}{1 + m_1 m_2} = \frac{\left| \frac{3}{5} + \frac{1}{3} \right|}{1 + \left( \frac{3}{5} \right) \left( -\frac{1}{3} \right)}$$

Simplifying the terms:

$$\tan \theta = \frac{\frac{14}{15}}{1 - \frac{1}{5}} = \frac{\frac{14}{15}}{\frac{4}{5}} = \frac{14}{12} = \frac{7}{6}$$

Now, to find $\theta$:

$$\theta = \tan^{-1} \left( \frac{7}{6} \right) \approx 49.4^\circ$$

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(b) Line 1 and Line 2:

Next, we calculate the slopes for the new set of lines:

1. Slope of $\ell_1$:
   • Points: $(3, -1)$ and $(5, 7)$
   • Using the slope formula:

$$m_1 = \frac{7 - (-1)}{5 - 3} = \frac{8}{2} = 4$$

2. Slope of $\ell_2$:
   • Points: $(2, 4)$ and $(-8, 2)$
   • Using the slope formula:

$$m_2 = \frac{2 - 4}{-8 - 2} = \frac{-2}{-10} = \frac{1}{5}$$

Now, we compute the angle:

$$\tan \theta = \frac{|m_1 - m_2|}{1 + m_1 m_2} = \frac{|4 - \frac{1}{5}|}{1 + 4 \cdot \frac{1}{5}}$$

Simplifying the terms:

$$\tan \theta = \frac{\frac{19}{15}}{\frac{9}{15}} = \frac{19}{9}$$

Now, to find $\theta$:

$$\theta = \tan^{-1} \left( \frac{19}{9} \right) \approx 64.65^\circ$$

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(c) Line 1 and Line 2:

For the final set of lines:

1. Slope of $\ell_1$:
   • Points: $(1, -7)$ and $(6, 4)$
   • Using the slope formula:

$$m_1 = \frac{4 - (-7)}{6 - 1} = \frac{11}{5} = \frac{3}{5}$$

2. Slope of $\ell_2$:
   • Points: $(-1, 2)$ and $(-6, -1)$
   • Using the slope formula:

$$m_2 = \frac{-1 - 2}{-6 - (-1)} = \frac{-3}{-5} = \frac{3}{5}$$

Here, we see that both slopes are equal, meaning the lines are parallel and there is no acute angle. Thus, the angle is:

$$\theta = \tan^{-1} (0) = 0^\circ$$

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(d) Line 1 and Line 2:

1. Slope of $\ell_1$:
   • Points: $(-9, -1)$ and $(3, -5)$
   • Using the slope formula:

$$m_1 = \frac{-5 - (-1)}{3 - (-9)} = \frac{-4}{12} = -\frac{1}{3}$$

2. Slope of $\ell_2$:
   • Points: $(2, 7)$ and $(-6, -7)$
   • Using the slope formula:

$$m_2 = \frac{-7 - 7}{-6 - 2} = \frac{-14}{-8} = \frac{7}{4}$$

Now, we calculate the angle:

$$\tan \theta = \frac{|m_1 - m_2|}{1 + m_1 m_2} = \frac{\left| -\frac{1}{3} - \frac{7}{4} \right|}{1 + \left( -\frac{1}{3} \right) \left( \frac{7}{4} \right)}$$

Simplifying the terms:

$$\tan \theta = \frac{\frac{25}{12}}{\frac{5}{12}} = 5$$

Now, to find $\theta$:

$$\theta = \tan^{-1} (5) \approx 78.69^\circ$$

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Key Formulas or Methods Used

• Slope Formula:
  $m = \frac{y_2 - y_1}{x_2 - x_1}$

• Angle Between Two Lines Formula:
  $\tan \theta = \frac{|m_1 - m_2|}{1 + m_1 m_2}$

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Summary of Steps

1. Calculate the slopes of both lines.
2. Apply the angle formula to find $\tan \theta$.
3. Find $\theta$ using the inverse tangent function.