The problem asks to find the angle between two lines given by the equations:

1. $d_1: 2x - 3y = 0$
2. $d_2: x + 3y - 7 = 0$

Let us solve it step by step.

Step 1: Rewrite the equations in slope-intercept form

The general equation of a straight line is: $Ax + By + C = 0$ The slope $m$ of the line is given by: $m = -\frac{A}{B}$

For $d_1: 2x - 3y = 0$:

• $A = 2$, $B = -3$ $m_1 = -\frac{2}{-3} = \frac{2}{3}$

For $d_2: x + 3y - 7 = 0$:

• $A = 1$, $B = 3$ $m_2 = -\frac{1}{3}$

Step 2: Formula for the angle between two lines

The angle $\theta$ between two lines is given by: $\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$

Substitute $m_1 = \frac{2}{3}$ and $m_2 = -\frac{1}{3}$: $\tan \theta = \left| \frac{-\frac{1}{3} - \frac{2}{3}}{1 + \frac{2}{3} \cdot -\frac{1}{3}} \right|$

Simplify the numerator: $-\frac{1}{3} - \frac{2}{3} = -\frac{3}{3} = -1$

Simplify the denominator: $1 + \frac{2}{3} \cdot -\frac{1}{3} = 1 - \frac{2}{9} = \frac{9}{9} - \frac{2}{9} = \frac{7}{9}$

Thus: $\tan \theta = \left| \frac{-1}{\frac{7}{9}} \right| = \left| -\frac{9}{7} \right| = \frac{9}{7}$

Step 3: Find the angle

The angle $\theta$ is: $\theta = \arctan\left(\frac{9}{7}\right)$

Using a calculator: $\theta \approx 52.13^\circ$

Final Answer:

The angle between the two lines is approximately: $\theta \approx 52.13^\circ$

Would you like a more detailed explanation or help with another problem? 😊

Related Questions:

1. What are the slopes of the lines and why are they important in calculating the angle?
2. How do we find the angle when the product of slopes is -1 (perpendicular lines)?
3. Can the angle between the lines be greater than $90^\circ$?
4. How does the calculation change if the lines are parallel?
5. Can we use dot product to find the angle between two lines?

Tip:

When working with angles between lines, always check the denominator in the formula for $\tan \theta$ to ensure it does not become zero (which would indicate parallel lines).