To find the [tex]$y$[/tex]-coordinate of the [tex]$y$[/tex]-intercept of line [tex]\( p \)[/tex], we need to use the slope-intercept form of a linear equation, which is given by:
[tex]\[ y = mx + b \][/tex]
Here, [tex]\( m \)[/tex] represents the slope, and [tex]\( b \)[/tex] represents the [tex]$y$[/tex]-intercept.
Given:
- The slope [tex]\( m \)[/tex] of line [tex]\( p \)[/tex] is [tex]\( -\frac{5}{3} \)[/tex].
- The [tex]$x$[/tex]-intercept of the line is at the point [tex]\((-6, 0)\)[/tex].
The [tex]$x$[/tex]-intercept is a point where the line crosses the [tex]$x$[/tex]-axis, meaning the [tex]$y$[/tex]-coordinate at that point is [tex]\( 0 \)[/tex]. We can use this point to find the [tex]$y$[/tex]-intercept by plugging it into the slope-intercept equation:
[tex]\[ y = mx + b \][/tex]
Given the point [tex]\((-6, 0)\)[/tex], we substitute [tex]\( x = -6 \)[/tex] and [tex]\( y = 0 \)[/tex] into the equation:
[tex]\[ 0 = \left( -\frac{5}{3} \right) (-6) + b \][/tex]
First, calculate the product of the slope and the [tex]\( x \)[/tex]-coordinate:
[tex]\[ \left( -\frac{5}{3} \right) (-6) = \frac{5}{3} \times 6 = \frac{30}{3} = 10 \][/tex]
Now our equation looks like:
[tex]\[ 0 = 10 + b \][/tex]
To find [tex]\( b \)[/tex], subtract [tex]\( 10 \)[/tex] from both sides:
[tex]\[ b = 0 - 10 \][/tex]
[tex]\[ b = -10 \][/tex]
So, the [tex]$y$[/tex]-coordinate of the [tex]$y$[/tex]-intercept of line [tex]\( p \)[/tex] is:
[tex]\[ \boxed{-10} \][/tex]
