Given:
Line P Q has points (-5, 3) and (5, 1).
Line R S has points (-4, -2) and (0, -4).
To find:
The relationship between lines PQ and RS.
Solution:
If a line passing through two points, then the slope of line is
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
Line P Q has points (-5, 3) and (5, 1). So, slope of line PQ is
[tex]m_1=\dfrac{1-3}{5-(-5)}[/tex]
[tex]m_1=\dfrac{-2}{5+5}[/tex]
[tex]m_1=\dfrac{-2}{10}[/tex]
[tex]m_1=\dfrac{-1}{5}[/tex]
Line R S has points (-4, -2) and (0, -4). So, slope of line RS is
[tex]m_2=\dfrac{-4-(-2)}{0-(-4)}[/tex]
[tex]m_2=\dfrac{-4+2}{0+4}[/tex]
[tex]m_2=\dfrac{-2}{4}[/tex]
[tex]m_2=\dfrac{-1}{2}[/tex]
Slopes of two parallel lines are equal.
[tex]m_1\neq m_2[/tex]
They are not parallel because their slopes are not equal.
Therefore, the correct option is C.
