Option B:
[tex]$\frac{1-(-1)}{2-0}=\frac{2-(-2)}{3-(-1)}[/tex]
Solution:
Triangle EFG and Triangle HJK are similar triangles.
Both triangles have the common line.
In triangle EFG, E(0, –1) and G(2, 1).
Here, [tex]x_1=0, y_1=-1, x_2=2, y_2=1[/tex]
Slope of the line:
[tex]$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
[tex]$m=\frac{1-(-1)}{2-0}[/tex] – – – – (1)
[tex]$m=\frac{2}{2}[/tex]
m = 1
In triangle HJK, H(–1, –2) and K(3, 2).
Here, [tex]x_1=-1, y_1=-2, x_2=3, y_2=2[/tex]
Slope of the line:
[tex]$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
[tex]$m=\frac{2-(-2)}{3-(-1)}[/tex] – – – – (2)
[tex]$m=\frac{4}{4}[/tex]
m = 1
Both slopes are equal. so that (1) and (2) are equal.
[tex]$\frac{1-(-1)}{2-0}=\frac{2-(-2)}{3-(-1)}[/tex]
Hence Option B is the correct answer.
