Let's solve this problem step-by-step:
1. Determine the slope of the line:
We have two points [tex]$(-4, -2)$[/tex] and [tex]$(2, 0)$[/tex]. The formula for the slope, [tex]\( m \)[/tex], between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
Plugging in the coordinates, we get:
[tex]\[
m = \frac{0 - (-2)}{2 - (-4)} = \frac{0 + 2}{2 + 4} = \frac{2}{6} = \frac{1}{3}
\][/tex]
So, the slope of the line is [tex]\( \frac{1}{3} \)[/tex].
2. Find the y-intercept of the line:
We can use the point-slope form of the line equation, [tex]\( y = mx + b \)[/tex], to determine the y-intercept [tex]\( b \)[/tex]. Substituting [tex]\( m = \frac{1}{3} \)[/tex] and one of the given points, say [tex]\((-4, -2)\)[/tex]:
[tex]\[
-2 = \frac{1}{3}(-4) + b
\][/tex]
Solving for [tex]\( b \)[/tex]:
[tex]\[
-2 = -\frac{4}{3} + b
\][/tex]
Add [tex]\(\frac{4}{3}\)[/tex] to both sides:
[tex]\[
-2 + \frac{4}{3} = b
\][/tex]
Convert [tex]\(-2\)[/tex] to a fraction with a common denominator of 3:
[tex]\[
-\frac{6}{3} + \frac{4}{3} = b
\][/tex]
This gives:
[tex]\[
b = -\frac{6}{3} + \frac{4}{3} = -\frac{2}{3}
\][/tex]
Thus, the y-intercept [tex]\( b \)[/tex] is [tex]\( -\frac{2}{3} \)[/tex].
3. Determine the value of [tex]\( a \)[/tex] such that the point [tex]\( (a, 1) \)[/tex] lies on the line:
We know the point [tex]\( (a, 1) \)[/tex] lies on the line, so we can use the line equation [tex]\( y = \frac{1}{3}x - \frac{2}{3} \)[/tex] and set [tex]\( y \)[/tex] to 1:
[tex]\[
1 = \frac{1}{3}a - \frac{2}{3}
\][/tex]
Add [tex]\(\frac{2}{3}\)[/tex] to both sides to isolate the term involving [tex]\( a \)[/tex]:
[tex]\[
1 + \frac{2}{3} = \frac{1}{3}a
\][/tex]
Convert 1 to a fraction with a common denominator of 3:
[tex]\[
\frac{3}{3} + \frac{2}{3} = \frac{1}{3}a
\][/tex]
Simplify:
[tex]\[
\frac{5}{3} = \frac{1}{3}a
\][/tex]
Multiply both sides by 3 to solve for [tex]\( a \)[/tex]:
[tex]\[
a = 5
\][/tex]
Therefore, the value of [tex]\( a \)[/tex] is [tex]\(\boxed{5}\)[/tex].
