Question

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

$3x + 4y = 18$
$3x - 4y = -18$
$3x + 4y = 6$
$3x - 4y = -6$

Explanation:

Step1: Find slope of given line

First, rewrite \(4x + 3y=9\) 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 the given line is \(m_1 =-\frac{4}{3}\).
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 point - slope form of a line is \(y - y_1=m(x - x_1)\), where \((x_1,y_1)=(-2,3)\) and \(m = \frac{3}{4}\).
\(y - 3=\frac{3}{4}(x + 2)\)
Multiply both sides by 4 to eliminate the fraction: \(4(y - 3)=3(x + 2)\)
Expand: \(4y-12 = 3x + 6\)
Rearrange to standard form \(Ax+By = C\): \(3x-4y=-18\) (subtract \(4y\) and add 12 to both sides: \(3x-4y=-18\))

We can also check by substituting \(x=-2\) and \(y = 3\) into each option:

• For \(3x + 4y=18\): \(3(-2)+4(3)=-6 + 12 = 6

eq18\)

• For \(3x-4y=-18\): \(3(-2)-4(3)=-6-12=-18\), which works.
• For \(3x + 4y=6\): \(3(-2)+4(3)=-6 + 12 = 6\), but the slope of this line is \(-\frac{3}{4}\), not \(\frac{3}{4}\) (since \(3x + 4y=6\) can be written as \(y=-\frac{3}{4}x+\frac{3}{2}\)), so it's not perpendicular.
• For \(3x-4y=-6\): \(3(-2)-4(3)=-6 - 12=-18

eq - 6\)

Answer:

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