Question

for problems 15 & 16, use the part of a city map shown.
15, which street is parallel to 1st ave? explain.
a. 2nd ave
b. main road
c. central ave
d. d street

1. a city planner wants to build a road perpendicular to d street. what is the slope of the new road?

show your work!

Explanation:

Step1: <Determine the slope of 1st Ave>

1st Ave is a vertical line. The slope of a vertical line is undefined. A vertical line has the equation \(x = a\) (where \(a\) is a constant). For a vertical line, the change in \(x\) (\(\Delta x\)) is \(0\) when calculating the slope \(m=\frac{\Delta y}{\Delta x}\). Since division by \(0\) is undefined, the slope of 1st Ave is undefined. A line parallel to a vertical line is also vertical.

Step2: <Identify the vertical street>

Looking at the options:

• 2nd Ave is a vertical line (has the form \(x = k\) where \(k\) is a constant).
• Main Road: Let's assume two points on Main Road. If we take two - point \((x_1,y_1)\) and \((x_2,y_2)\). For example, if we assume Main Road passes through \((- 4,-2)\) and \((0, - 1)\), then \(m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{-1+2}{0 + 4}=\frac{1}{4}\) (not vertical).
• Central Ave: Assume two points. If it passes through \((-2,4)\) and \((0,3)\), then \(m=\frac{3 - 4}{0+2}=-\frac{1}{2}\) (not vertical).
• D Street: It is a horizontal line (has the form \(y = c\) where \(c\) is a constant), and the slope of a horizontal line \(m = 0\) (since \(\Delta y=0\) and \(m=\frac{\Delta y}{\Delta x}=0\) for \(x_1

eq x_2\)).

Step3: <Find the slope of D Street>

D Street is a horizontal line. Let's assume two points on D Street, say \((x_1,y_1)\) and \((x_2,y_2)\) where \(y_1=y_2\). The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Since \(y_2 - y_1 = 0\) (for \(x_1
eq x_2\)), \(m = 0\).
If a line is perpendicular to a horizontal line (slope \(m_1 = 0\)), and we know that if \(m_1\) and \(m_2\) are slopes of two perpendicular lines, then \(m_1\times m_2=- 1\). Let \(m_1 = 0\), then the slope of the line perpendicular to it is undefined (vertical line). But if we consider the general case of a non - vertical/non - horizontal line. Let's assume two points on Central Ave: say \((x_1,y_1)=(-2,4)\) and \((x_2,y_2)=(0,3)\), \(m_{Central}=\frac{3 - 4}{0 + 2}=-\frac{1}{2}\). If a line is perpendicular to a line with slope \(m\), the slope of the perpendicular line is \(m'=-\frac{1}{m}\). For a horizontal line \(m = 0\), we can think in terms of geometric properties. A horizontal line and a vertical line are perpendicular. Since D Street is horizontal (slope \(m = 0\)), a line perpendicular to D Street is vertical (undefined slope). But if we use the formula for non - horizontal/non - vertical lines in a wrong way (as a limiting case). Let's assume we use the concept of the product of slopes of perpendicular lines \(m_1\times m_2=-1\). If \(m_1 = 0\) (D Street), and we consider the limit as \(m_1\to0\) for \(m_2=-\frac{1}{m_1}\), \(m_2\to\infty\) (undefined, vertical line).

Answer:

A. 2nd Ave