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Name the quadrilateral formed, if any, by the following points, and give reasons for your answers:$A (-1, -2), B (1, 0), C (-1, 2), D (-3, 0)$
Given:
Given points are $A (-1, -2), B (1, 0), C (-1, 2), D (-3, 0)$.
To do:
We have to find the quadrilateral formed, if any, by the given points.
Solution:
We know that,
The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).
Therefore,
\( \mathrm{AB}=\sqrt{(1+1)^{2}+(0+2)^{2}} \)
Squaring on both sides, we get,
\( \mathrm{AB}^{2}=(1+1)^{2}+(0+2)^{2} \)
\( =(2)^{2}+(2)^{2} \)
\( =4+4 \)
\( =8 \)
\( \mathrm{BC}^{2}=(-1-1)^{2}+(2-0)^{2} \)
\( =(-2)^{2}+(2)^{2} \)
\( =4+4 \)
\( =8 \)
\( \mathrm{CD}^{2}=(-3+1)^{2}+(0-2)^{2} \)
\( =(-2)^{2}+(-2)^{2} \)
\( =4+4 \)
\( =8 \)
\( \mathrm{DA}^{2}=(-1+3)^{2}+(-2+0)^{2} \)
\( =(2)^{2}+(-2)^{2} \)
\( =4+4 \)
\( =8 \)
\( \mathrm{AC}^{2}=(-1+1)^{2}+(2+2)^{2} \)
\( =(0)^{2}+(4)^{2} \)
\( =0+16 \)
\( =16 \)
\( \mathrm{BD}^{2}=(-3-1)^{2}+(0-0)^{2} \)
\( =(-4)^{2}+0 \)
\( =16 \)
Here,
\( AB^2=BC^2=CD^2=DA^2 \) and \( AC^2=BD^2 \)
This implies,
\( AB=BC=CD=DA \) and \( AC=BD \)
All the sides are equal and the diagonals are equal to each other.
Therefore, the quadrilateral formed by the points \( A, B, C, D \) is a square.
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