Question

1. which equation represents the line that is perpendicular to the graph of $4x + 3y = 9$ and passes through the point $(-2, 3)$?\

\\(\circ\\) $3x + 4y = 18$\
\\(\circ\\) $3x + 4y = 6$\
\\(\circ\\) $3x - 4y = -18$\
\\(\circ\\) $3x - 4y = -6$

Explanation:

Step1: Find slope of given line

The given line is \(4x + 3y=9\). Rewrite in slope - intercept form \(y = mx + b\) (where \(m\) is the slope):
\(3y=-4x + 9\), so \(y=-\frac{4}{3}x + 3\). The slope of this line, \(m_1=-\frac{4}{3}\).

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 \(-\frac{4}{3}\times m_2=-1\), so \(m_2=\frac{3}{4}\).

Step2: Use point - slope form

The line passes through \((-2,3)\) and has slope \(m_2 = \frac{3}{4}\). The point - slope form of a line is \(y - y_1=m(x - x_1)\) (where \((x_1,y_1)=(-2,3)\) and \(m = \frac{3}{4}\)):
\(y - 3=\frac{3}{4}(x + 2)\)

Step3: Convert to standard form

Multiply both sides by 4 to get rid of the fraction: \(4(y - 3)=3(x + 2)\)
Expand: \(4y-12 = 3x+6\)
Rearrange to standard form \(Ax+By = C\): \(3x-4y=-18\)

Answer:

\(3x - 4y=-18\) (the option with \(3x - 4y=-18\))