Question

1. select all the equations that are perpendicular to the graph of 6x + 18y = 5. a. y = 3x - 10 b. x = 3 c. y + 6 = 3(x - 15) d. 3x + 9y = 8 e. 2x - 3y = 5

Explanation:

Step1: Find slope of given line

First, rewrite \(6x + 18y=5\) in slope - intercept form \(y = mx + b\) (where \(m\) is the slope).
Subtract \(6x\) from both sides: \(18y=-6x + 5\).
Divide by 18: \(y=\frac{-6}{18}x+\frac{5}{18}=\frac{-1}{3}x+\frac{5}{18}\). So the slope of the given line, \(m_1=-\frac{1}{3}\).

Step2: Recall perpendicular slope rule

If two lines are perpendicular, the product of their slopes \(m_1\times m_2=- 1\). Let the slope of the perpendicular line be \(m_2\). Then \(-\frac{1}{3}\times m_2=-1\), so \(m_2 = 3\).

Step3: Analyze each option

• Option A: \(y = 3x-10\). The slope \(m = 3\). Check \(m_1\times m=-\frac{1}{3}\times3=-1\). Perpendicular.
• Option B: \(x = 3\) is a vertical line (undefined slope). The given line has slope \(-\frac{1}{3}\) (non - vertical). Vertical and non - vertical lines are perpendicular? No, because the product of undefined and \(-\frac{1}{3}\) is not \(-1\) (undefined).
• Option C: Simplify \(y + 6=3(x - 15)\). Expand: \(y+6 = 3x-45\). Then \(y=3x-51\). Slope \(m = 3\). \(m_1\times m=-\frac{1}{3}\times3=-1\). Perpendicular.
• Option D: Rewrite \(3x + 9y=8\) as \(9y=-3x + 8\), \(y=\frac{-3}{9}x+\frac{8}{9}=\frac{-1}{3}x+\frac{8}{9}\). Slope \(m=-\frac{1}{3}\), same as given line (parallel, not perpendicular).
• Option E: Rewrite \(2x-3y = 5\) as \(-3y=-2x + 5\), \(y=\frac{2}{3}x-\frac{5}{3}\). Slope \(m=\frac{2}{3}\). \(m_1\times m=-\frac{1}{3}\times\frac{2}{3}=-\frac{2}{9}

eq - 1\). Not perpendicular.

Answer:

A. \(y = 3x-10\), C. \(y + 6=3(x - 15)\)