Answer :
To determine which of the given equations could represent the central street [tex]\(PQ\)[/tex] parallel or perpendicular to the street passing through points [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we'll start by identifying the slope of the line given by [tex]\(-7x + 3y = -21.5\)[/tex].
First, let's rearrange the equation [tex]\(-7x + 3y = -21.5\)[/tex] into the slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] represents the slope:
[tex]\[ -7x + 3y = -21.5 \][/tex]
Add [tex]\(7x\)[/tex] to both sides:
[tex]\[ 3y = 7x - 21.5 \][/tex]
Divide both sides by [tex]\(3\)[/tex]:
[tex]\[ y = \frac{7}{3}x - \frac{21.5}{3} \][/tex]
From this, we observe that the slope [tex]\(m\)[/tex] of the line passing through points [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is [tex]\(\frac{7}{3}\)[/tex].
Next, we'll analyze the slopes of the equations provided in the options, and we aim to identify a line that's either parallel (same slope) or perpendicular (negative reciprocal slope) to the given line.
### Option A: [tex]\(-3x + 4y = 3\)[/tex]
Rearrange it into slope-intercept form:
[tex]\[ -3x + 4y = 3 \][/tex]
Add [tex]\(3x\)[/tex] to both sides:
[tex]\[ 4y = 3x + 3 \][/tex]
Divide both sides by [tex]\(4\)[/tex]:
[tex]\[ y = \frac{3}{4}x + \frac{3}{4} \][/tex]
The slope is [tex]\(\frac{3}{4}\)[/tex].
### Option B: [tex]\(3x + 7y = 63\)[/tex]
Rearrange it into slope-intercept form:
[tex]\[ 3x + 7y = 63 \][/tex]
Subtract [tex]\(3x\)[/tex] from both sides:
[tex]\[ 7y = -3x + 63 \][/tex]
Divide both sides by [tex]\(7\)[/tex]:
[tex]\[ y = -\frac{3}{7}x + 9 \][/tex]
The slope is [tex]\(-\frac{3}{7}\)[/tex].
### Option C: [tex]\(2x + y = 20\)[/tex]
Rearrange it into slope-intercept form:
[tex]\[ 2x + y = 20 \][/tex]
Subtract [tex]\(2x\)[/tex] from both sides:
[tex]\[ y = -2x + 20 \][/tex]
The slope is [tex]\(-2\)[/tex].
### Option D: [tex]\(7x + 3y = 70\)[/tex]
Rearrange it into slope-intercept form:
[tex]\[ 7x + 3y = 70 \][/tex]
Subtract [tex]\(7x\)[/tex] from both sides:
[tex]\[ 3y = -7x + 70 \][/tex]
Divide both sides by [tex]\(3\)[/tex]:
[tex]\[ y = -\frac{7}{3}x + \frac{70}{3} \][/tex]
The slope is [tex]\(-\frac{7}{3}\)[/tex].
Having identified the slopes of the given lines, we look for a line with either the same slope as [tex]\(\frac{7}{3}\)[/tex] (parallel) or the slope [tex]\(-\frac{3}{7}\)[/tex] (perpendicular).
Comparing:
- Option A: Slope [tex]\(\frac{3}{4}\)[/tex]
- Option B: Slope [tex]\(-\frac{3}{7}\)[/tex]
- Option C: Slope [tex]\(-2\)[/tex]
- Option D: Slope [tex]\(-\frac{7}{3}\)[/tex]
We see that the slope in option B [tex]\(-\frac{3}{7}\)[/tex] is the negative reciprocal of [tex]\(\frac{7}{3}\)[/tex] which indicates that it is perpendicular to the given line, fitting the problem's scenario.
Thus, the equation of the central street [tex]\(PQ\)[/tex] is:
[tex]\[ \boxed{3x + 7y = 63} \][/tex]
First, let's rearrange the equation [tex]\(-7x + 3y = -21.5\)[/tex] into the slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] represents the slope:
[tex]\[ -7x + 3y = -21.5 \][/tex]
Add [tex]\(7x\)[/tex] to both sides:
[tex]\[ 3y = 7x - 21.5 \][/tex]
Divide both sides by [tex]\(3\)[/tex]:
[tex]\[ y = \frac{7}{3}x - \frac{21.5}{3} \][/tex]
From this, we observe that the slope [tex]\(m\)[/tex] of the line passing through points [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is [tex]\(\frac{7}{3}\)[/tex].
Next, we'll analyze the slopes of the equations provided in the options, and we aim to identify a line that's either parallel (same slope) or perpendicular (negative reciprocal slope) to the given line.
### Option A: [tex]\(-3x + 4y = 3\)[/tex]
Rearrange it into slope-intercept form:
[tex]\[ -3x + 4y = 3 \][/tex]
Add [tex]\(3x\)[/tex] to both sides:
[tex]\[ 4y = 3x + 3 \][/tex]
Divide both sides by [tex]\(4\)[/tex]:
[tex]\[ y = \frac{3}{4}x + \frac{3}{4} \][/tex]
The slope is [tex]\(\frac{3}{4}\)[/tex].
### Option B: [tex]\(3x + 7y = 63\)[/tex]
Rearrange it into slope-intercept form:
[tex]\[ 3x + 7y = 63 \][/tex]
Subtract [tex]\(3x\)[/tex] from both sides:
[tex]\[ 7y = -3x + 63 \][/tex]
Divide both sides by [tex]\(7\)[/tex]:
[tex]\[ y = -\frac{3}{7}x + 9 \][/tex]
The slope is [tex]\(-\frac{3}{7}\)[/tex].
### Option C: [tex]\(2x + y = 20\)[/tex]
Rearrange it into slope-intercept form:
[tex]\[ 2x + y = 20 \][/tex]
Subtract [tex]\(2x\)[/tex] from both sides:
[tex]\[ y = -2x + 20 \][/tex]
The slope is [tex]\(-2\)[/tex].
### Option D: [tex]\(7x + 3y = 70\)[/tex]
Rearrange it into slope-intercept form:
[tex]\[ 7x + 3y = 70 \][/tex]
Subtract [tex]\(7x\)[/tex] from both sides:
[tex]\[ 3y = -7x + 70 \][/tex]
Divide both sides by [tex]\(3\)[/tex]:
[tex]\[ y = -\frac{7}{3}x + \frac{70}{3} \][/tex]
The slope is [tex]\(-\frac{7}{3}\)[/tex].
Having identified the slopes of the given lines, we look for a line with either the same slope as [tex]\(\frac{7}{3}\)[/tex] (parallel) or the slope [tex]\(-\frac{3}{7}\)[/tex] (perpendicular).
Comparing:
- Option A: Slope [tex]\(\frac{3}{4}\)[/tex]
- Option B: Slope [tex]\(-\frac{3}{7}\)[/tex]
- Option C: Slope [tex]\(-2\)[/tex]
- Option D: Slope [tex]\(-\frac{7}{3}\)[/tex]
We see that the slope in option B [tex]\(-\frac{3}{7}\)[/tex] is the negative reciprocal of [tex]\(\frac{7}{3}\)[/tex] which indicates that it is perpendicular to the given line, fitting the problem's scenario.
Thus, the equation of the central street [tex]\(PQ\)[/tex] is:
[tex]\[ \boxed{3x + 7y = 63} \][/tex]
