To find the equation of the central street [tex]\( PQ \)[/tex], we need to consider that it could either be parallel or perpendicular to the given lane passing through points [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
The given equation for the lane is:
[tex]\[ -7x + 3y = -21.5 \][/tex]
1. Determine the slope of the line through [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
- Convert the given equation to the slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope.
[tex]\[ -7x + 3y = -21.5 \][/tex]
[tex]\[ 3y = 7x - 21.5 \][/tex]
[tex]\[ y = \frac{7}{3}x - \frac{21.5}{3} \][/tex]
2. Now, we can see that the slope ([tex]\( m \)[/tex]) of the line [tex]\( AB \)[/tex] is [tex]\( \frac{7}{3} \)[/tex].
2. Identify the possible slopes for the central street [tex]\( PQ \)[/tex]:
- For a line parallel to [tex]\( AB \)[/tex]:
- The slope remains the same: [tex]\( \frac{7}{3} = 2.3333333333333335 \)[/tex].
- For a line perpendicular to [tex]\( AB \)[/tex]:
- The slope is the negative reciprocal of [tex]\( \frac{7}{3} \)[/tex]:
[tex]\[ -\frac{1}{\frac{7}{3}} = -\frac{3}{7} = -0.42857142857142855 \][/tex]
Therefore, the slopes for the central street [tex]\( PQ \)[/tex] are either [tex]\( 2.3333333333333335 \)[/tex] (if parallel) or [tex]\( -0.42857142857142855 \)[/tex] (if perpendicular).
