Answer :
To solve the problem of finding the equation of the central street [tex]\(PQ\)[/tex] which is perpendicular to the lane passing through [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we need to determine the slope relationships between these lines. Let's break down the solution step-by-step:
### Step 1: Identify the slope of the given line
The equation of the lane passing through [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is given as:
[tex]\[ -7x + 3y = -21.5 \][/tex]
First, we need to rewrite this equation in the slope-intercept form [tex]\( y = mx + c \)[/tex], where [tex]\(m\)[/tex] is the slope.
Rearrange the equation to solve for [tex]\(y\)[/tex]:
[tex]\[ 3y = 7x - 21.5 \][/tex]
[tex]\[ y = \frac{7}{3}x - \frac{21.5}{3} \][/tex]
From this, we see that the slope [tex]\(m\)[/tex] of the line is:
[tex]\[ m = \frac{7}{3} \][/tex]
### Step 2: Determine the slope of the perpendicular line
Lines that are perpendicular to each other have slopes that are negative reciprocals. Thus, the slope of any line perpendicular to the given line will be the negative reciprocal of [tex]\(\frac{7}{3}\)[/tex].
Calculate the negative reciprocal:
[tex]\[ m_{\text{perpendicular}} = -\frac{1}{\left(\frac{7}{3}\right)} = -\frac{3}{7} \][/tex]
### Step 3: Check the options for the perpendicular slope
We need to identify which of the given options has a slope of [tex]\(-\frac{3}{7}\)[/tex]. Each provided equation must be converted to the slope-intercept form [tex]\( y = mx + c \)[/tex] so that we can identify its slope.
#### Option A: [tex]\(-3x + 4y = 3\)[/tex]
Rearrange to solve for [tex]\(y\)[/tex]:
[tex]\[ 4y = 3x + 3 \][/tex]
[tex]\[ y = \frac{3}{4}x + \frac{3}{4} \][/tex]
The slope here is [tex]\(\frac{3}{4}\)[/tex], which is not [tex]\(-\frac{3}{7}\)[/tex].
#### Option B: [tex]\(3x + 7y = 63\)[/tex]
Rearrange to solve for [tex]\(y\)[/tex]:
[tex]\[ 7y = -3x + 63 \][/tex]
[tex]\[ y = -\frac{3}{7}x + 9 \][/tex]
The slope here is [tex]\(-\frac{3}{7}\)[/tex], which matches the required slope for the perpendicular line.
#### Option C: [tex]\(2x + y = 20\)[/tex]
Rearrange to solve for [tex]\(y\)[/tex]:
[tex]\[ y = -2x + 20 \][/tex]
The slope here is [tex]\(-2\)[/tex], which is not [tex]\(-\frac{3}{7}\)[/tex].
#### Option D: [tex]\(7x + 3y = 70\)[/tex]
Rearrange to solve for [tex]\(y\)[/tex]:
[tex]\[ 3y = -7x + 70 \][/tex]
[tex]\[ y = -\frac{7}{3}x + \frac{70}{3} \][/tex]
The slope here is [tex]\(-\frac{7}{3}\)[/tex], which is not [tex]\(-\frac{3}{7}\)[/tex].
### Conclusion
Among the provided options, only Option B: [tex]\(3x + 7y = 63\)[/tex] has a slope of [tex]\(-\frac{3}{7}\)[/tex], making it the correct equation for the perpendicular street [tex]\(PQ\)[/tex].
Therefore, the equation of the central street [tex]\(PQ\)[/tex] is:
[tex]\[ \boxed{3x + 7y = 63} \][/tex]
### Step 1: Identify the slope of the given line
The equation of the lane passing through [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is given as:
[tex]\[ -7x + 3y = -21.5 \][/tex]
First, we need to rewrite this equation in the slope-intercept form [tex]\( y = mx + c \)[/tex], where [tex]\(m\)[/tex] is the slope.
Rearrange the equation to solve for [tex]\(y\)[/tex]:
[tex]\[ 3y = 7x - 21.5 \][/tex]
[tex]\[ y = \frac{7}{3}x - \frac{21.5}{3} \][/tex]
From this, we see that the slope [tex]\(m\)[/tex] of the line is:
[tex]\[ m = \frac{7}{3} \][/tex]
### Step 2: Determine the slope of the perpendicular line
Lines that are perpendicular to each other have slopes that are negative reciprocals. Thus, the slope of any line perpendicular to the given line will be the negative reciprocal of [tex]\(\frac{7}{3}\)[/tex].
Calculate the negative reciprocal:
[tex]\[ m_{\text{perpendicular}} = -\frac{1}{\left(\frac{7}{3}\right)} = -\frac{3}{7} \][/tex]
### Step 3: Check the options for the perpendicular slope
We need to identify which of the given options has a slope of [tex]\(-\frac{3}{7}\)[/tex]. Each provided equation must be converted to the slope-intercept form [tex]\( y = mx + c \)[/tex] so that we can identify its slope.
#### Option A: [tex]\(-3x + 4y = 3\)[/tex]
Rearrange to solve for [tex]\(y\)[/tex]:
[tex]\[ 4y = 3x + 3 \][/tex]
[tex]\[ y = \frac{3}{4}x + \frac{3}{4} \][/tex]
The slope here is [tex]\(\frac{3}{4}\)[/tex], which is not [tex]\(-\frac{3}{7}\)[/tex].
#### Option B: [tex]\(3x + 7y = 63\)[/tex]
Rearrange to solve for [tex]\(y\)[/tex]:
[tex]\[ 7y = -3x + 63 \][/tex]
[tex]\[ y = -\frac{3}{7}x + 9 \][/tex]
The slope here is [tex]\(-\frac{3}{7}\)[/tex], which matches the required slope for the perpendicular line.
#### Option C: [tex]\(2x + y = 20\)[/tex]
Rearrange to solve for [tex]\(y\)[/tex]:
[tex]\[ y = -2x + 20 \][/tex]
The slope here is [tex]\(-2\)[/tex], which is not [tex]\(-\frac{3}{7}\)[/tex].
#### Option D: [tex]\(7x + 3y = 70\)[/tex]
Rearrange to solve for [tex]\(y\)[/tex]:
[tex]\[ 3y = -7x + 70 \][/tex]
[tex]\[ y = -\frac{7}{3}x + \frac{70}{3} \][/tex]
The slope here is [tex]\(-\frac{7}{3}\)[/tex], which is not [tex]\(-\frac{3}{7}\)[/tex].
### Conclusion
Among the provided options, only Option B: [tex]\(3x + 7y = 63\)[/tex] has a slope of [tex]\(-\frac{3}{7}\)[/tex], making it the correct equation for the perpendicular street [tex]\(PQ\)[/tex].
Therefore, the equation of the central street [tex]\(PQ\)[/tex] is:
[tex]\[ \boxed{3x + 7y = 63} \][/tex]
