Given:
DE contains the points D(1, -2) and E(3, 4).
FG contains the points F(-1, 2) and G(4, 0).
To find:
Is DE perpendicular to FG.
Solution:
Slope of DE:
[tex]$m=\frac{y_2-y_1}{x_2-x_1}[/tex]
Here [tex]x_1=1, y_1=-2, x_2=3, y_2=4[/tex]
[tex]$m=\frac{4-(-2)}{3-1}[/tex]
[tex]$m=\frac{6}{2}[/tex]
m = 3
Slope of FG:
[tex]$m=\frac{y_2-y_1}{x_2-x_1}[/tex]
Here [tex]x_1=-1, y_1=2, x_2=4, y_2=0[/tex]
[tex]$m=\frac{0-2}{4-(-1)}[/tex]
[tex]$m=\frac{-2}{4+1}[/tex]
[tex]$m=\frac{-2}{5}[/tex]
Two lines are perpendicular if product their slopes are -1.
Slope of DE × Slope of FG
[tex]$=3\times \frac{-2}{5}[/tex]
[tex]$= \frac{-6}{5}[/tex]
≠ -1
The solution is no, because the product of the slopes is not -1.
Answer:
The second option: no, because the product of the slopes is not -1.
