To find the standard form of the equation of the line passing through point [tex]\( B(-10, -5) \)[/tex] and perpendicular to the line given by [tex]\( 6x - 5y = 24 \)[/tex], follow these steps:
### Step 1: Identify the slope of the given line
First, we need to rewrite the equation [tex]\( 6x - 5y = 24 \)[/tex] in the slope-intercept form [tex]\( y = mx + b \)[/tex].
1. Start with the given equation:
[tex]\[
6x - 5y = 24
\][/tex]
2. Rearrange the equation to solve for [tex]\( y \)[/tex]:
[tex]\[
-5y = -6x + 24
\][/tex]
[tex]\[
y = \frac{6}{5}x - \frac{24}{5}
\][/tex]
Thus, the slope [tex]\( m_1 \)[/tex] of the given line is [tex]\( \frac{6}{5} \)[/tex].
### Step 2: Determine the slope of the perpendicular line
The slope of the line perpendicular to another is the negative reciprocal of the other line's slope. Therefore, if the slope of the given line is [tex]\( m_1 = \frac{6}{5} \)[/tex], then the slope [tex]\( m_2 \)[/tex] of the perpendicular line is:
[tex]\[
m_2 = -\frac{1}{m_1} = -\frac{1}{\frac{6}{5}} = -\frac{5}{6}
\][/tex]
### Step 3: Use the point-slope form to find the equation of the perpendicular line
Now, we will use the point-slope form of the line equation, which is:
[tex]\[
y - y_1 = m(x - x_1)
\][/tex]
where [tex]\( m \)[/tex] is the slope of the line, and [tex]\((x_1, y_1)\)[/tex] is a point on the line [tex]\((x_1 = -10, y_1 = -5)\)[/tex].
1. Substitute [tex]\((x_1, y_1)\)[/tex] and [tex]\( m_2 = -\frac{5}{6} \)[/tex] into the point-slope form:
[tex]\[
y - (-5) = -\frac{5}{6}(x - (-10))
\][/tex]
[tex]\[
y + 5 = -\frac{5}{6}(x + 10)
\][/tex]
2. Simplify the equation:
[tex]\[
y + 5 = -\frac{5}{6}x - \frac{5}{6} \cdot 10
\][/tex]
[tex]\[
y + 5 = -\frac{5}{6}x - \frac{50}{6}
\][/tex]
[tex]\[
y + 5 = -\frac{5}{6}x - \frac{25}{3}
\][/tex]
3. Finally, combine like terms to solve for [tex]\( y \)[/tex]:
[tex]\[
y = -\frac{5}{6}x - \frac{25}{3} - 5
\][/tex]
[tex]\[
y = -\frac{5}{6}x - \frac{25}{3} - \frac{15}{3}
\][/tex]
[tex]\[
y = -\frac{5}{6}x - \frac{40}{3}
\][/tex]
### Step 4: Convert the equation to standard form [tex]\( Ax + By = C \)[/tex]
To change this to standard form, multiply through by 6 (the denominator of [tex]\(\frac{5}{6}\)[/tex]) to clear the fraction:
[tex]\[
6y = -5x - 80
\][/tex]
[tex]\[
5x + 6y = -80
\][/tex]
Thus, the standard form of the equation of the line that passes through [tex]\( B(-10, -5) \)[/tex] and is perpendicular to the graph of [tex]\( 6x - 5y = 24 \)[/tex] is:
[tex]\[
5x + 6y = -80
\][/tex]
