Answer:
Option (c)
Step-by-step explanation:
Slope of a line that passing through two points M(-1, 4) and N(2, -5),
[tex]m_1[/tex] = [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]
= [tex]\frac{-1-2}{4+5}[/tex]
= -([tex]\frac{3}{9}[/tex])
= -[tex]\frac{1}{3}[/tex]
To find the line perpendicular line to MN we will use the property,
[tex]m_1\times m_2=-1[/tex]
Where [tex]m_1[/tex] and [tex]m_2[/tex] are the slopes of two perpendicular lines.
Slope of line perpendicular to MN [tex](m_2)[/tex] will be,
[tex]-\frac{1}{3}\times m_2=-1[/tex]
[tex]m_2=3[/tex]
Slope of line joining two points J(-3, -4) and K(3, -2),
Slope = [tex]\frac{-4+2}{-3-3}=\frac{1}{3}[/tex]
Slope of line joining two points A(-3, 2) and B(3, 0)
Slope = [tex]\frac{2-0}{-3-3}=-\frac{1}{3}[/tex]
Slope of the line joining points E(0, -3) and F(2, 3),
Slope = [tex]\frac{-3-3}{0-2}[/tex] = 3
Therefore, line EF is perpendicular to the line MN.
Option (c) is the answer.
