the length of the line segment \( P Q \). 6 The line \( l \) is parallel to the line \( x+2 y=6 \). Given that \( l \) passes through the point \( (3,4) \), find the coordinates of the point where \( l \) meets the \( x \)-axis.

To find the length of the line segment \( PQ \), we first need to determine the equation of the line \( l \) that is parallel to the line \( x + 2y = 6 \). The slope of the given line can be found by rewriting it in slope-intercept form: 1. Rewrite \( x + 2y = 6 \) as \( 2y = -x + 6 \) or \( y = -\frac{1}{2}x + 3 \). 2. Therefore, the slope is \( -\frac{1}{2} \). Since line \( l \) is parallel, it has the same slope. Using the point-slope form of the line equation at the point \( (3, 4) \), we get: \[ y - 4 = -\frac{1}{2}(x - 3) \] Simplifying gives: \[ y - 4 = -\frac{1}{2}x + \frac{3}{2} \] \[ y = -\frac{1}{2}x + \frac{3}{2} + 4 \] \[ y = -\frac{1}{2}x + \frac{11}{2} \] To find the intersection with the \( x \)-axis, set \( y = 0 \): \[ 0 = -\frac{1}{2}x + \frac{11}{2} \] \[ \frac{1}{2}x = \frac{11}{2} \] \[ x = 11 \] Thus, the coordinates where line \( l \) meets the \( x \)-axis are \( (11, 0) \). The length of the segment \( PQ \) depends on the specific points \( P \) and \( Q \) given in your original question, but the crucial calculation for this line segment is complete!