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Name the quadrilateral formed, if any, by the following points, and give reasons for your answers:$A (-3, 5), B (3, 1), C (0, 3), D (-1, -4)$
Given:
Given points are $A (-3, 5), B (3, 1), C (0, 3), D (-1, -4)$.
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,
\( \mathrm{AB}=\sqrt{(3+3)^{2}+(1-5)^{2}} \)
Squaring on both sides, we get,
\( \mathrm{AB}^{2}=(3+3)^{2}+(1-5)^{2} \)
\( =(6)^{2}+(-4)^{2} \)
\( =36+16 \)
\( =52 \)
\( \mathrm{BC}^{2}=(0-3)^{2}+(3-1)^{2} \)
\( =(-3)^{2}+(2)^{2} \)
\( =9+4 \)
\( =13 \)
\( \mathrm{CD}^{2}=(-1-0)^{2}+(-4-3)^{2} \)
\( =(-1)^{2}+(-7)^{2} \)
\( =1+49 \)
\( =50 \)
\( \mathrm{DA}^{2}=(-3+1)^{2}+(5+4)^{2} \)
\( =(-2)^{2}+(9)^{2} \)
\( =4+81 \)
\( =85 \)
\( \mathrm{AC}^{2}=(0+3)^{2}+(3-5)^{2} \)
\( =(3)^{2}+ (-2)^{2} \)
\( =9+4 \)
\( =13 \)
In \( \Delta \mathrm{ABC} \),
\( \mathrm{AB}=\sqrt{52}, \mathrm{AC}=\sqrt{13}, \mathrm{BC}=\sqrt{13} \)
\( \mathrm{AC}+\mathrm{BC}=\sqrt{13}+\sqrt{13}=2 \sqrt{13} \)
\( \mathrm{AB}=\sqrt{52}=2\sqrt{13} \)
\(\Rightarrow \mathrm{AC}+\mathrm{BC}=\mathrm{AB} \)
This implies, the points $A, B ,C$ are collinear.
Therefore, \( \Delta \mathrm{ABC} \) is not possible.
Hence, \( \mathrm{ABCD} \) is not a quadrilateral.
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