To determine which lines have a slope of [tex]\(-\frac{4}{5}\)[/tex] and contain the point [tex]\((0, 1)\)[/tex], follow these steps:
1. Examine the slopes: Identify the lines and their slopes. In this scenario, [tex]$\overleftrightarrow{BC}$[/tex] and [tex]$\overleftrightarrow{CE}$[/tex] both have slopes of [tex]\(-\frac{4}{5}\)[/tex].
2. Check the point containment: Verify if the point [tex]\((0, 1)\)[/tex] lies on each of these lines.
- For a line with slope [tex]\(-\frac{4}{5}\)[/tex] to contain a point like [tex]\((0, 1)\)[/tex], it must satisfy the line equation in the slope-intercept form [tex]\(y = mx + b\)[/tex]. Here [tex]\(m\)[/tex] is the slope and [tex]\(b\)[/tex] is the y-intercept.
- Given slopes are [tex]\(-\frac{4}{5}\)[/tex]; we substitute them into the generic line equation.
3. Determine the equation of the lines:
- Starting with [tex]$\overleftrightarrow{BC}$[/tex], let the equation be [tex]\(y = -\frac{4}{5}x + b\)[/tex]. If the point [tex]\((0, 1)\)[/tex] lies on the line:
[tex]\[
1 = -\frac{4}{5}(0) + b \implies b = 1
\][/tex]
Thus, the equation is [tex]\(y = -\frac{4}{5}x + 1\)[/tex].
- For [tex]$\overleftrightarrow{CE}$[/tex], we follow the same method:
[tex]\[
1 = -\frac{4}{5}(0) + b \implies b = 1
\][/tex]
The equation is also [tex]\(y = -\frac{4}{5}x + 1\)[/tex].
4. Conclude: Both lines, [tex]$\overleftrightarrow{BC}$[/tex] and [tex]$\overleftrightarrow{CE}$[/tex], have the same slope of [tex]\(-\frac{4}{5}\)[/tex] and both go through the point [tex]\((0, 1)\)[/tex].
Thus, the lines [tex]\(\overleftrightarrow{BC}\)[/tex] and [tex]\(\overleftrightarrow{CE}\)[/tex] both satisfy the conditions of having the slope [tex]\(-\frac{4}{5}\)[/tex] and containing the point [tex]\((0, 1)\)[/tex].
Hence, the answer is: both [tex]\(\overleftrightarrow{BC}\)[/tex] and [tex]\(\overleftrightarrow{CE}\)[/tex].
