Given:
AB formed by (-2,13) and (0,3).
CD formed by (-5,0) and (10,3).
To find:
Whether the segments AB and CD are parallel, perpendicular, or neither.
Solution:
Slope formula:
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
AB formed by (-2,13) and (0,3). So, the slope of AB is:
[tex]m_1=\dfrac{3-13}{0-(-2)}[/tex]
[tex]m_1=\dfrac{-10}{2}[/tex]
[tex]m_1=-5[/tex]
CD formed by (-5,0) and (10,3). So, slope of CD is:
[tex]m_2=\dfrac{3-0}{10-(-5)}[/tex]
[tex]m_2=\dfrac{3}{10+5}[/tex]
[tex]m_2=\dfrac{3}{15}[/tex]
[tex]m_2=\dfrac{1}{5}[/tex]
Since [tex]m_1\neq m_2[/tex], therefore the segments AB and CD are not parallel.
[tex]m_1\times m_2=-5\times \dfrac{1}{5}[/tex]
[tex]m_1\times m_2=-1[/tex]
Since [tex]m_1\times m_2=-1[/tex], therefore the segments AB and CD are perpendicular because product of slopes of two perpendicular lines is always -1.
Hence, the segments AB and CD are perpendicular.
Answer:
AB is perpendicular to CD.
Step-by-step explanation:
AB formed by (-2, 13) and (0, 3)
CD formed by (-5, 0) and (10, 3)
Slope of a line passing through two points is
[tex]m= \frac{y''-y'}{x''- x'}[/tex]
The slope of line AB is
[tex]m= \frac{3- 13}{0+2} = -5[/tex]
The slope of line CD is
[tex]m'= \frac{3 -0 }{10+5} = \frac{1}{5}[/tex]
As the product of m and m' is -1 so the lines AB and CD are perpendicular to each other.