Question

1. select all equations whose graphs are lines perpendicular to the graph of 3x + 2y = 6. a. 3x − 2y = 4 b. 2x + 3y = 6 c. 2x − 3y = 8 d. (y − 4) = \frac{2}{3}(x − 6) e. (y − 2) = -\frac{3}{2}(x − 8) f. y = \frac{2}{3}x g. y = \frac{3}{2}x + 3 keep trying. (select all that apply.)

Explanation:

Step1: Find slope of \(3x + 2y = 6\)

Rewrite in slope - intercept form \(y=mx + b\) (where \(m\) is the slope).
\(2y=-3x + 6\), so \(y=-\frac{3}{2}x+3\). The slope of the given line, \(m_1 =-\frac{3}{2}\).

Step2: Determine slope of perpendicular lines

If two lines with slopes \(m_1\) and \(m_2\) are perpendicular, then \(m_1\times m_2=- 1\). Let the slope of the perpendicular line be \(m_2\).
We have \(-\frac{3}{2}\times m_2=-1\), solving for \(m_2\) gives \(m_2=\frac{2}{3}\).

Step3: Find slopes of each option

• Option A: \(3x - 2y = 4\)

Rewrite as \(2y = 3x-4\), \(y=\frac{3}{2}x - 2\). Slope \(m=\frac{3}{2}
eq\frac{2}{3}\).

• Option B: \(2x + 3y = 6\)

Rewrite as \(3y=-2x + 6\), \(y=-\frac{2}{3}x + 2\). Slope \(m =-\frac{2}{3}
eq\frac{2}{3}\).

• Option C: \(2x - 3y = 8\)

Rewrite as \(3y=2x - 8\), \(y=\frac{2}{3}x-\frac{8}{3}\). Slope \(m=\frac{2}{3}\).

• Option D: \((y - 4)=\frac{2}{3}(x - 6)\)

This is in point - slope form \(y - y_1=m(x - x_1)\), slope \(m=\frac{2}{3}\).

• Option E: \((y - 2)=-\frac{3}{2}(x - 8)\)

Slope \(m =-\frac{3}{2}
eq\frac{2}{3}\).

• Option F: \(y=\frac{2}{3}x\)

Slope \(m=\frac{2}{3}\).

• Option G: \(y=\frac{3}{2}x + 3\)

Slope \(m=\frac{3}{2}
eq\frac{2}{3}\).

Answer:

C. \(2x - 3y = 8\), D. \((y - 4)=\frac{2}{3}(x - 6)\), F. \(y=\frac{2}{3}x\)