Question

a private plane and a jetliner both fly over a park that is at (-2, 2) on a map. the path of the jetliner follows the line ( y = \frac{1}{2}x + 3 ). the private plane flies perpendicular to the jetliner. what is the equation in slope - intercept form of the line that represents the private plane’s flight path?
( y = square x + square )

Explanation:

Step1: Find slope of jetliner's line

The jetliner's line is \( y = \frac{1}{2}x + 3 \). The slope (\( m_1 \)) is \( \frac{1}{2} \). For perpendicular lines, the product of slopes is -1. Let the private plane's slope be \( m_2 \). So, \( \frac{1}{2} \times m_2 = -1 \), which gives \( m_2 = -2 \).

Step2: Use point - slope form

The private plane passes through \((-2, 2)\) and has slope -2. The point - slope form is \( y - y_1 = m(x - x_1) \), where \( (x_1,y_1)=(-2,2) \) and \( m = -2 \). Substituting, we get \( y - 2 = -2(x - (-2)) \), which simplifies to \( y - 2 = -2(x + 2) \).

Step3: Convert to slope - intercept form

Expand \( y - 2 = -2x - 4 \). Then add 2 to both sides: \( y = -2x - 4 + 2 \), so \( y = -2x - 2 \).

Answer:

\( y = -2x - 2 \) (So in the boxes, the first box is -2 and the second box is -2)