To solve the problem, we need to find out which of the provided options (A, B, C, or D) represents the equation of the central street PQ given that it must have a unique slope relationship (either parallel or perpendicular) with the given lane passing through points [tex]\(A\)[/tex] and [tex]\(B\)[/tex], represented by the equation [tex]\(-7x + 3y = -21.5\)[/tex].
1. Convert the given equation to slope-intercept form to determine the slope:
The given equation is:
[tex]\[
-7x + 3y = -21.5
\][/tex]
Rewrite it in the form [tex]\( y = mx + b \)[/tex] (slope-intercept form) by solving for [tex]\(y\)[/tex]:
[tex]\[
3y = 7x - 21.5
\][/tex]
[tex]\[
y = \frac{7}{3}x - \frac{21.5}{3}
\][/tex]
So, the slope [tex]\( m \)[/tex] of the given line is:
[tex]\[
m = \frac{7}{3}
\][/tex]
2. Determine the properties needed for a parallel or perpendicular line:
- For a line to be parallel to the given line, it must have the same slope [tex]\( \frac{7}{3} \)[/tex].
- For a line to be perpendicular to the given line, the product of their slopes should be [tex]\(-1\)[/tex]. Here, if [tex]\( m_1 = \frac{7}{3} \)[/tex], then the slope [tex]\( m_2 \)[/tex] of the perpendicular line should satisfy:
[tex]\[
\frac{7}{3} \times m_2 = -1 \implies m_2 = -\frac{3}{7}
\][/tex]
3. Check each option to see which satisfies either property:
A. [tex]\(-3x + 4y = 3\)[/tex]:
Convert to slope-intercept form:
[tex]\[
4y = 3x + 3
\][/tex]
[tex]\[
y = \frac{3}{4}x + \frac{3}{4}
\][/tex]
The slope is [tex]\( \frac{3}{4} \)[/tex], which does not match either [tex]\( \frac{7}{3} \)[/tex] or [tex]\(-\frac{3}{7} \)[/tex].
B. [tex]\(3x + 7y = 63\)[/tex]:
Convert to slope-intercept form:
[tex]\[
7y = -3x + 63
\][/tex]
[tex]\[
y = -\frac{3}{7}x + 9
\][/tex]
The slope is [tex]\(-\frac{3}{7} \)[/tex], which is exactly the perpendicular slope to [tex]\( \frac{7}{3} \)[/tex].
C. [tex]\(2x + y = 20\)[/tex]:
Convert to slope-intercept form:
[tex]\[
y = -2x + 20
\][/tex]
The slope is [tex]\(-2 \)[/tex], which does not match either [tex]\( \frac{7}{3} \)[/tex] or [tex]\(-\frac{3}{7} \)[/tex].
D. [tex]\(7x + 3y = 70\)[/tex]:
Convert to slope-intercept form:
[tex]\[
3y = -7x + 70
\][/tex]
[tex]\[
y = -\frac{7}{3}x + \frac{70}{3}
\][/tex]
The slope is [tex]\(-\frac{7}{3} \)[/tex], which is not perpendicular to [tex]\( \frac{7}{3} \)[/tex] but its negative reciprocal which doesn’t follow the problem’s required relationship.
Among all the options, the equation that satisfies the condition of being perpendicular to the given line is B:
[tex]\[
3x + 7y = 63
\][/tex]
