Question

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

Answer 1

Step-by-step explanation:

which is required ans of this question

Answer 2

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

Step-by-step explanation:

Given,

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

Using distance formula,

$\sqrt{(x_{2}-x_{1} )^{2}+(y_{2}-y_{1} )^{2}}$

∴ AB = $\sqrt{(1-5)^2+(5-6)^2}$

= $\sqrt{(-4)^2+(-1)^2} =\sqrt{16+1} =\sqrt{17}$

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

$=\sqrt{(1)^2+(-4)^2}=\sqrt{1+16}=\sqrt{17}$

CD = $\sqrt{(6-2)^2+(2-1)^2}=\sqrt{(4)^2+(1)^2}=\sqrt{16+1}=\sqrt{17}$

DA  $=\sqrt{(6-5)^2+(2-6)^2}\\=\sqrt{(1)^2+(-4)^2}=\sqrt{1+16}=\sqrt{17}$

∴ AB = BC = CD = DA = $\sqrt{17}$, all sides are equa.

Diagonals,

AC = $\sqrt{(2-5)^2+(6-1)^2}=\sqrt{(-3)^2+(5)^2}=\sqrt{9+25} =\sqrt{34}$

BD $= \sqrt{(6-1)^2+(2-5)^2}=\sqrt{(5)^2+(-3)^2}=\sqrt{25+9} =\sqrt{34}$

∴ Diagonal AC = Diagonal BD $=\sqrt{34}$

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