Given
A Quadrilateral ABCD with points A (2, -1) , B (4, -4), C (2, -7), D (0,-4).
To find the equations of line symmetry in quadrilateral ABCD.
Explanation:
It is given that,
From the figure, it is clear that the quadrilateral is a Rhombus.
Since the Rhombus has two lines of symmetry AC and BD.
Therefore,
The equation of the line AC is,
[tex]\begin{gathered} \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1} \\ \frac{y-(-1)}{-7-(-1)}=\frac{x-2}{2-2} \\ \frac{y+1}{-7+1}=\frac{x-2}{0} \\ \frac{y+1}{6}=\frac{x-2}{0} \\ x-2=0 \\ x=2 \end{gathered}[/tex]
Therefore the equation of the line of symmetry AC is, x=2.
Similarly, the equation of the line of symmetry BD is, y= -4, because BD is a horizontal line.
Reasoning:
Since the lines of symmetry are horizontal line and vertical line.
Hence, x=2, y=-4 are the equations of the lines of symmetry.
