(i) Show that the points A(5, 6), B(1, 5), C(2, 1) and D(6,2) are the vertices of a square.

Question:

(i) Show that the points A(5, 6), B(1, 5), C(2, 1) and D(6,2) are the vertices of a square.

(ii) Prove hat the points A (2, 3) B(−2,2) C(−1,−2), and D(3, −1) are the vertices of a square ABCD.

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

In a square all the sides are equal to each other. And also the diagonals are also equal to each other.

Here the four points are A(5,6), B(1,5), C(2,1) and D(6,2).

First let us check if all the four sides are equal.

$A B=\sqrt{(5-1)^{2}+(6-5)^{2}}$

$=\sqrt{(4)^{2}+(1)^{2}}$

 

$=\sqrt{16+1}$

$A B=\sqrt{17}$

$B C=\sqrt{(1-2)^{2}+(5-1)^{2}}$

$=\sqrt{(-1)^{2}+(4)^{2}}$

 

$=\sqrt{1+16}$

$B C=\sqrt{17}$

$C D=\sqrt{(2-6)^{2}+(1-2)^{2}}$

$=\sqrt{(-4)^{2}+(-1)^{2}}$

 

$=\sqrt{16+1}$

$C D=\sqrt{17}$

$A D=\sqrt{(5-6)^{2}+(6-2)^{2}}$

$=\sqrt{(-1)^{2}+(4)^{2}}$

 

$=\sqrt{1+16}$

$A D=\sqrt{17}$

Here, we see that all the sides are equal, so it has to be a rhombus.

Now let us find out the lengths of the diagonals of this rhombus.

$A C=\sqrt{(5-2)^{2}+(6-1)^{2}}$

$=\sqrt{(3)^{2}+(5)^{2}}$

 

$=\sqrt{9+25}$

$A C=\sqrt{34}$

$B D=\sqrt{(1-6)^{2}+(5-2)^{2}}$

$=\sqrt{(-5)^{2}+(3)^{2}}$

 

$=\sqrt{25+9}$

$B D=\sqrt{34}$

Now since the diagonals of the rhombus are also equal to each other this rhombus has to be a square.

Hence we have proved that the quadrilateral formed by the given four points is a.

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