Prove that $(4, 3), (6, 4), (5, 6)$ and $(3, 5)$ are the angular points of a square.

Given:

Given vertices are $(4, 3), (6, 4), (5, 6)$ and $(3, 5)$.

To do:

We have to prove that the points $(4, 3), (6, 4), (5, 6)$ and $(3, 5)$ are the vertices of a square.

Solution:

Let the vertices of the square be $A(4, 3), B(6, 4), C(5, 6)$ and $D(3, 5)$.

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,

\( \mathrm{AB}=\sqrt{(6-4)^{2}+(4-3)^{2}} \)

\( =\sqrt{(2)^{2}+(1)^{2}} \)

\( =\sqrt{4+1} \)

\( =\sqrt{5} \)

\( \mathrm{BC}=\sqrt{(5-6)^{2}+(6-4)^{2}} \)

\( =\sqrt{(-1)^{2}+(2)^{2}} \)

\( =\sqrt{1+4} \)

\( =\sqrt{5} \)

\( \mathrm{CD}=\sqrt{(3-5)^{2}+(5-6)^{2}} \)

\( =\sqrt{(-2)^{2}+(-1)^{2}} \)

\( =\sqrt{4+1} \)

\( =\sqrt{5} \)

\( \mathrm{DA}=\sqrt{(4-3)^{2}+(3-5)^{2}} \)

\( =\sqrt{(1)^{2}+(-2)^{2}} \)

\( =\sqrt{1+4} \)

\( =\sqrt{5} \)

\( \mathrm{AC}=\sqrt{(5-4)^{2}+(6-3)^{2}} \)

\( =\sqrt{(1)^{2}+(3)^{2}} \)

\( =\sqrt{1+9} \)

\( =\sqrt{10} \)

\( \mathrm{BD}=\sqrt{(3-6)^{2}+(5-4)^{2}} \)

\( =\sqrt{(-3)^{2}+(1)^{2}} \)

\( =\sqrt{9+1} \)

\( =\sqrt{10} \)

Here,

\( \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA} \)

\( \mathrm{AC}=\mathrm{BD} \)

Therefore, $(4, 3), (6, 4), (5, 6)$ and $(3, 5)$ are the vertices of a square.

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Updated on: 10-Oct-2022

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