Given:
Line a passes through (2, 10) and (4, 13).
Line b passes through (4, 9) and (6, 12).
Line c passes through (2, 10) and (4, 9).
To find:
Which of the lines, if any are perpendicular.
Solution:
If a line passes through two points, then the slope of line is
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
Line a passes through (2, 10) and (4, 13). So, slope of this line is
[tex]m_a=\dfrac{13-10}{4-2}=\dfrac{3}{2}[/tex]
Line b passes through (4, 9) and (6, 12). So, slope of this line is
[tex]m_b=\dfrac{12-9}{6-4}=\dfrac{3}{2}[/tex]
Line c passes through (2, 10) and (4,9). So, slope of this line is
[tex]m_c=\dfrac{9-10}{4-2}=\dfrac{-1}{2}[/tex]
Product of slopes of to perpendicular lines is -1.
[tex]m_a\cdot m_b=\dfrac{3}{2}\times \dfrac{3}{2}=\dfrac{9}{4}\neq -1[/tex]
[tex]m_b\cdot m_c=\dfrac{3}{2}\times \dfrac{-1}{2}=\dfrac{-3}{4}\neq -1[/tex]
[tex]m_a\cdot m_c=\dfrac{3}{2}\times \dfrac{-1}{2}=\dfrac{-3}{4}\neq -1[/tex]
Therefore, any of these lines are not perpendicular to each other.
