Answer :
To determine the equation of the central street [tex]\(PQ\)[/tex] that is perpendicular to the lane passing through points [tex]\(A\)[/tex] and [tex]\(B\)[/tex], we need a clear understanding of the relationship between perpendicular lines.
First, let’s start by analyzing the given equation of the lane passing through [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ -7x + 3y = -21.5 \][/tex]
To find the slope of this line, we should convert it to the slope-intercept form, [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope.
Starting with the given equation:
[tex]\[ -7x + 3y = -21.5 \][/tex]
Resolve 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 line [tex]\(AB\)[/tex] is:
[tex]\[ m = \frac{7}{3} \][/tex]
For two lines to be perpendicular, the product of their slopes must be [tex]\(-1\)[/tex]. Thus, the slope of the perpendicular line [tex]\(PQ\)[/tex] will be the negative reciprocal of [tex]\(\frac{7}{3}\)[/tex]:
[tex]\[ m_{PQ} = -\frac{1}{\left(\frac{7}{3}\right)} = -\frac{3}{7} \][/tex]
Now, we need to identify which of the given equations has this slope [tex]\(-\frac{3}{7}\)[/tex]. Let’s rewrite each of the given equations in slope-intercept form and identify their slopes.
A. [tex]\(-3x + 4y = 3\)[/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].
B. [tex]\(3x + 7y = 63\)[/tex]
[tex]\[ 7y = -3x + 63 \][/tex]
[tex]\[ y = -\frac{3}{7}x + 9 \][/tex]
The slope here is [tex]\(-\frac{3}{7}\)[/tex].
C. [tex]\(2x + y = 20\)[/tex]
[tex]\[ y = -2x + 20 \][/tex]
The slope here is [tex]\(-2\)[/tex].
D. [tex]\(7x + 3y = 70\)[/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].
Among the given choices, only option B has the slope [tex]\(-\frac{3}{7}\)[/tex].
So, the equation of the central street [tex]\(PQ\)[/tex] is:
[tex]\[ 3x + 7y = 63 \][/tex]
First, let’s start by analyzing the given equation of the lane passing through [tex]\(A\)[/tex] and [tex]\(B\)[/tex]:
[tex]\[ -7x + 3y = -21.5 \][/tex]
To find the slope of this line, we should convert it to the slope-intercept form, [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope.
Starting with the given equation:
[tex]\[ -7x + 3y = -21.5 \][/tex]
Resolve 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 line [tex]\(AB\)[/tex] is:
[tex]\[ m = \frac{7}{3} \][/tex]
For two lines to be perpendicular, the product of their slopes must be [tex]\(-1\)[/tex]. Thus, the slope of the perpendicular line [tex]\(PQ\)[/tex] will be the negative reciprocal of [tex]\(\frac{7}{3}\)[/tex]:
[tex]\[ m_{PQ} = -\frac{1}{\left(\frac{7}{3}\right)} = -\frac{3}{7} \][/tex]
Now, we need to identify which of the given equations has this slope [tex]\(-\frac{3}{7}\)[/tex]. Let’s rewrite each of the given equations in slope-intercept form and identify their slopes.
A. [tex]\(-3x + 4y = 3\)[/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].
B. [tex]\(3x + 7y = 63\)[/tex]
[tex]\[ 7y = -3x + 63 \][/tex]
[tex]\[ y = -\frac{3}{7}x + 9 \][/tex]
The slope here is [tex]\(-\frac{3}{7}\)[/tex].
C. [tex]\(2x + y = 20\)[/tex]
[tex]\[ y = -2x + 20 \][/tex]
The slope here is [tex]\(-2\)[/tex].
D. [tex]\(7x + 3y = 70\)[/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].
Among the given choices, only option B has the slope [tex]\(-\frac{3}{7}\)[/tex].
So, the equation of the central street [tex]\(PQ\)[/tex] is:
[tex]\[ 3x + 7y = 63 \][/tex]
