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Name the quadrilateral formed, if any, by the following points, and give reasons for your answers:$A (4, 5), B (7, 6), C (4, 3), D (1, 2)$
Given:
Given points are $A (4, 5), B (7, 6), C (4, 3), D (1, 2)$.
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,
\( \mathrm{AB}=\sqrt{(7-4)^{2}+(6-5)^{2}} \)
Squaring on both sides, we get,
\( \mathrm{AB}^{2}=(7-4)^{2}+(6-5)^{2} \)
\( =(3)^{2}+(1)^{2} \)
\( =9+1 \)
\( =10 \)
\( B C^{2}=(4-7)^{2}+(3-6)^{2} \)
\( =(3)^{2}+(-3)^{2} \)
\( =9+9 \)
\( =18 \)
\( \mathrm{CD}^{2}=(1-4)^{2}+(2-3)^{2} \)
\( =(-3)^{2}+(-1)^{2} \)
\( =9+1 \)
\( =10 \)
\( \mathrm{DA}^{2}=(4-1)^{2}+(5-2)^{2} \)
\( =(3)^{2}+(3)^{2} \)
\( =9+9 \)
\( =18 \)
\( \mathrm{AC}^{2}=(4-4)^{2}+(3-5)^{2} \)
\( =(0)^{2}+(-2)^{2} \)
\( =0+4 \)
\( =4 \)
\( \mathrm{BD}^{2}=(1-7)^{2}+(2-6)^{2} \)
\( =(-6)^{2}+(-4)^{2} \)
\( =36+16 \)
\( =52 \)
Here,
\( AB^2=CD^2 \), \( BC^2=DA^2 \) and \( AC^2≠BD^2 \)
This implies,
\( A B=C D \), \( B C=D A \) and \( AC≠BD \)
Opposite sides are equal and diagonals are not equal.
Therefore, the quadrilateral formed by the points \( A, B, C, D \) is a parallelogram.
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