Question

If the line \(l_1 : 3y - 2x = 3\) is the angular bisector of the lines \(l_2 : x - y + 1 = 0\) and \(l_3\) : \(αx + βy + 17 = 0\), then \(α^2 + β^2  - α - β\) is equal to__

Answer 1

Correct Answer: 348

We are tasked with analyzing the intersection and verification of lines. The given lines are:

\( \ell_1 : 3y - 2x = 3, \quad \ell_2 : x - y + 1 = 0. \)

Step 1: Find the point of intersection

The point of intersection of \( \ell_1 \) and \( \ell_2 \) is:

\( P \equiv (0, 1). \)

Step 2: Verify the line \( \ell_3 \)

The point \( P(0, 1) \) lies on the line:

\( \ell_3 : \alpha x - \beta y + 17 = 0. \)

Substitute \( P(0, 1) \) into \( \ell_3 \):

\( \alpha(0) - \beta(1) + 17 = 0 \implies \beta = -17. \)

Step 3: Find the image of a random point about \( \ell_2 \)

Consider a random point \( Q \equiv (-1, 0) \) on \( \ell_2 : x - y + 1 = 0 \). The image of \( Q(-1, 0) \) about \( \ell_2 \) is:

\( Q' \equiv \left( -\frac{17}{13}, \frac{6}{13} \right). \)

This is calculated using the formula for the reflection of a point about a line.

Step 4: Verify that \( Q' \) lies on \( \ell_3 \)

Substitute \( Q' \left( -\frac{17}{13}, \frac{6}{13} \right) \) into \( \ell_3 \):

Substitute \( \beta = -17 \):

\( \ell_3 : \alpha x - \beta y + 17 = 0. \)

\( \alpha \left( -\frac{17}{13} \right) - (-17) \left( \frac{6}{13} \right) + 17 = 0. \)

Simplify:

\( -\frac{17\alpha}{13} + \frac{102}{13} + 17 = 0. \)

Equating coefficients, solve for \( \alpha \):

\( \alpha = 7. \)

Step 5: Final verification

Now substitute \( \alpha = 7 \) and \( \beta = -17 \) into the condition:

\( \alpha^2 + \beta^2 - \alpha - \beta = 348. \)