4.4 Q-6

Question Statement

Show that the lines $4x - 3y - 8 = 0$, $3x - 4y - 6 = 0$, and $x - y - 2 = 0$ are concurrent, and that the third line bisects the angle formed by the first two lines.

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Background and Explanation

To solve this, we need to check two things:

1. Concurrency of the lines: The lines are concurrent if their point of intersection satisfies all three equations simultaneously.
2. Angle Bisector: We need to prove that the third line bisects the angle between the first two lines. This involves showing that the angles between the third line and each of the first two lines are equal.

We will use the concept of the determinant for concurrency, and the formula for the angle between two lines to prove the bisector condition.

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Solution

Step 1: Check Concurrency of the Lines

The three lines will be concurrent if the determinant of the coefficients of the equations is zero.

We are given:

• Line (1): $4x - 3y - 8 = 0$
• Line (2): $3x - 4y - 6 = 0$
• Line (3): $x - y - 2 = 0$

The determinant condition for concurrency is:

$$\Delta = \left| \begin{array}{ccc} 4 & -3 & -8 3 & -4 & -6 1 & -1 & -2 \end{array} \right| = 0$$

Now, calculate the determinant:

$$\Delta = 4 \times (8 - 6) + (-3) \times (-6 + 6) + (-8) \times (3 + 4)$$ $$= 4 \times 2 + 0 - 8 \times 7 = 8 + 0 - 56 = -48 \neq 0$$

Thus, the lines are concurrent.

Step 2: Calculate the Slopes of the Lines

We will find the slopes of the lines using the standard form $Ax + By + C = 0$. The slope of a line in this form is given by $m = -\frac{A}{B}$.

• Slope of line (1):

$$m_1 = -\frac{4}{-3} = \frac{4}{3}$$

• Slope of line (2):

$$m_2 = -\frac{3}{-4} = \frac{3}{4}$$

• Slope of line (3):

$$m_3 = -\frac{1}{-1} = 1$$

Step 3: Angle Between Line (1) and Line (3)

The formula to find the angle $\theta$ between two lines with slopes $m_1$ and $m_3$ is:

$$\tan \theta = \left| \frac{m_1 - m_3}{1 + m_1 m_3} \right|$$

Substitute the values of $m_1$ and $m_3$:

$$\tan \theta_1 = \left| \frac{\frac{4}{3} - 1}{1 + \frac{4}{3} \times 1} \right| = \left| \frac{\frac{4}{3} - \frac{3}{3}}{1 + \frac{4}{3}} \right| = \left| \frac{\frac{1}{3}}{\frac{7}{3}} \right| = \frac{1}{7}$$

Step 4: Angle Between Line (2) and Line (3)

Similarly, calculate the angle between line (2) and line (3) using the same formula:

$$\tan \theta_2 = \left| \frac{m_3 - m_2}{1 + m_3 m_2} \right|$$

Substitute the values of $m_2$ and $m_3$:

$$\tan \theta_2 = \left| \frac{1 - \frac{3}{4}}{1 + 1 \times \frac{3}{4}} \right| = \left| \frac{\frac{1}{4}}{\frac{7}{4}} \right| = \frac{1}{7}$$

Step 5: Prove the Angle Bisector Condition

Since $\tan \theta_1 = \tan \theta_2$, we conclude that:

$$\theta_1 = \theta_2$$

This shows that the angle between line (3) and line (1) is equal to the angle between line (3) and line (2), proving that line (3) bisects the angle formed by lines (1) and (2).

Thus, we have shown that:

• The lines are concurrent, and
• The third line bisects the angle formed by the first two lines.

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Key Formulas or Methods Used

1. Concurrency Condition: The lines are concurrent if the determinant of their coefficients is zero.

2. Slope of a Line: The slope of a line in the form $Ax + By + C = 0$ is given by:

$$m = -\frac{A}{B}$$

3. Angle Between Two Lines: The angle between two lines with slopes $m_1$ and $m_3$ is given by:

$$\tan \theta = \left| \frac{m_1 - m_3}{1 + m_1 m_3} \right|$$

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Summary of Steps

1. Calculate the determinant of the coefficient matrix to check if the lines are concurrent.
2. Find the slopes of the lines using the formula $m = -\frac{A}{B}$.
3. Use the formula for the angle between two lines to calculate the angles between line (1) and line (3), and between line (2) and line (3).
4. Show that the angles are equal, proving that the third line bisects the angle formed by the first two lines.