Answer with Step-by-step explanation:
We are given that a triangle ABC. In triangle ABC , the coordinates of A is at (0,0), Point B is at [tex](x^2,0)[/tex], Point C is at [tex](x_1,y_1)[/tex], Point D is at [tex] (\frac{x-1}{2},\frac{y-1}{2})[/tex], Point E is at [tex](\frac{x-1\frac{x-2}{2},\frac{y-1}{2})[/tex].
The coordinates of point E can be write as [tex](\frac{3x-4}{2}.\frac{y-1}{2})[/tex].
Slope formula of a line passing through two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex]is given by
=[tex]\frac{y_2-y_1}{x_2-x_1}[/tex].
By using slope formula of a line
Slope of line AB which passing through the points A (0,0) and B[tex](x^2,0)[/tex]
Where [tex]x_1=0,x_2=x^2,y_1=0,y_2=0[/tex]
=[tex]\frac{0-0}{x^2-0}[/tex]
Slope of line AB which passing through the points A(0,0) and B[tex](x^2,0),m_1[/tex]=0
Slope of line DE which passing through the points D[tex](\frac{x-1}{2},\frac{y-1}{2})[/tex] and E[tex](\frac{3x-4}{2},\frac{y-1}{2})[/tex]
Where[tex] x_1=\frac{x-1}{2},x_2=\frac{3x-4}{2},y_1=\frac{y-1}{2}, y_2=\frac{y-1}{2}[/tex]
=[tex]\frac{\frac{y-1}{2}-\frac{y-1}{2}}{\frac{3x-4}{2}-\frac{x-1}{2}}[/tex]
Slope of line DE,[tex]m_2=\frac{\frac{y-1-y+1}{2}}{\frac{3x-4-x+1}{2}}[/tex]
Slope of line DE,[tex]m_2=\frac{0}{\frac{2x-3}{2}}[/tex]
Slope of line DE which passing through the points D[tex](\frac{x-1}{2},\frac{y-1}{2})[/tex], and E[tex](\frac{3x-4}{2},\frac{y-1}{2})[/tex]
=0
We know that when two lines are parallel then thier slopes are equal.
[tex]m_1=m_2[/tex]
Slope of line DE = Slope of line AB
Hence, the line DE and line AB are parallel lines.
