The equation of line containing the angle bisector of the lines \(3 x-4 y-2=0\) and \(5 x-12 y+2=0\) is ....... (a) \(7 x+4 y-18=0\) (b) \(4 x-7 y-1=0\) (c) \(4 \mathrm{x}-7 \mathrm{y}+1=0\) (d) None of these

To find the slopes of the lines, rewrite both of them in the slope-intercept form, y = mx + b, where m is the slope. Original equations: \(3x - 4y - 2 = 0\) \(5x - 12y + 2 = 0\) Lines in slope-intercept form: \(y = \frac{3}{4}x - \frac{1}{2}\) (Line 1) \(y = \frac{5}{12}x - \frac{1}{6}\) (Line 2) Slopes: \(m_1 = \frac{3}{4}\) \(m_2 = \frac{5}{12}\)

Now, we will find the tangent of the half-angle, using the following formula: \(tan(\frac{\phi}{2}) = \frac{m_2 - m_1}{1 + m_1 m_2}\) Plug in the slopes from Step 1: \(tan(\frac{\phi}{2}) = \frac{\frac{5}{12} - \frac{3}{4}}{1 + (\frac{3}{4})(\frac{5}{12})}\) Simplify the expression: \(tan(\frac{\phi}{2}) = \frac{\frac{-1}{3}}{\frac{25}{16}} = -\frac{16}{75}\) Since the tangent of the half-angle is negative, we know that one of the angle bisectors will have a positive slope, and the other will have a negative slope. At this point, we can observe which of the given answer choices have lines with slopes of opposite signs. (a) 7x + 4y - 18 = 0 → y = -\(\frac{7}{4}\)x + \(\frac{9}{2}\) → negative slope (b) 4x - 7y - 1 = 0 → y = \(\frac{4}{7}\)x - \(\frac{1}{7}\) → positive slope (c) 4x - 7y + 1 = 0 → y = \(\frac{4}{7}\)x + \(\frac{1}{7}\) → positive slope From the elimination above, we can conclude that the angle bisector lies between lines (b) and (c). To determine which one is correct, observe the numerator of the tangent expression and match the signs to the slopes of the given lines.

To match the signs, compare the signs of the differences in the slopes of the given lines and the tangent expression (numerator). Since our tangent of the half-angle is negative, we are looking for a pair of lines where the difference in their slopes is negative. \(m_b = \frac{4}{7}\), \(m_c = \frac{4}{7}\) (slope of lines (b) and (c)) \(m_1 = \frac{3}{4}\) (slope of Line 1) Differences: \(m_b - m_1 = \frac{4}{7} - \frac{3}{4} = \frac{1}{28}\) → positive \(m_c - m_1 = \frac{4}{7} - \frac{3}{4} = \frac{1}{28}\) → positive Since both differences are positive, the angle bisector must be between Line 1 and the reflection of either (b) or (c). The reflection of a line with a positive slope will make the slope negative. Reflect the given line equations (b) and (c): Reflected lines: (b') y = -\(\frac{4}{7}\)x + \(r_1\) (c') y = -\(\frac{4}{7}\)x + \(r_2\) Now, let's calculate the differences again between Line 1 and the reflected lines (b') and (c'): Differences: \(m_{b'} - m_1 = -\frac{4}{7} - \frac{3}{4} = -\frac{25}{28}\) → negative \(m_{c'} - m_1 = -\frac{4}{7} - \frac{3}{4} = -\frac{25}{28}\) → negative Now both differences are negative, and the angle bisector must be between Line 1 and the reflection of either (b') or (c'). Note that -\(\frac{25}{28}\) is a negative version of the denominator from the tangent expression so it represents the correct angle bisector. From this, we can conclude that the equation of the line containing the angle bisector is:

(b) \(4x - 7y - 1 = 0\)