Question
Prove that (4, 3), (6, 4) (5, 6) and (3, 5) are the angular points of a square.
✓
Answer
Let the given points be, A(4, 3), B(6, 4) C(5, 6) and D(3, 5) respectively. Then, 
Now, $\text{AB}=\sqrt{(6-4)^2+(4-3)^2}$ $=\sqrt{4+1}$ $=\sqrt{5}\ \text{units}$ and $\text{BC}=\sqrt{(5-6)^2+(6-4)^2}$$=\sqrt{1+4}$
$=\sqrt{5}\ \text{units}$ $\text{CD}=\sqrt{(3-5)^2+(5-6)^2}$ $=\sqrt{4+1}$ $=\sqrt{5}\ \text{units}$ $\text{AD}=\sqrt{(3-4)^2+(5-3)^2}$ $=\sqrt{1+4}$ $=\sqrt{5}\ \text{units}$ Thus, AB = BC = CD = AD Diagonal $\text{AC}=\sqrt{(5-4)^2+(6-3)^2}$ $=\sqrt{1+9}$ $=\sqrt{10}\ \text{units}$ Diagonal $\text{BD}=\sqrt{(3-6)^2+(5-4)^2}$ $=\sqrt{9+1}$ $=\sqrt{10}\ \text{units}$ $\therefore$ AB = BC = CD = AD and diagonal AC = diagonal BD. Thus, ABCD is a quadrilateral in which all sides are equal and the diagonal are equal. Hence, ABCD is a square.
Now, $\text{AB}=\sqrt{(6-4)^2+(4-3)^2}$ $=\sqrt{4+1}$ $=\sqrt{5}\ \text{units}$ and $\text{BC}=\sqrt{(5-6)^2+(6-4)^2}$$=\sqrt{1+4}$$=\sqrt{5}\ \text{units}$ $\text{CD}=\sqrt{(3-5)^2+(5-6)^2}$ $=\sqrt{4+1}$ $=\sqrt{5}\ \text{units}$ $\text{AD}=\sqrt{(3-4)^2+(5-3)^2}$ $=\sqrt{1+4}$ $=\sqrt{5}\ \text{units}$ Thus, AB = BC = CD = AD Diagonal $\text{AC}=\sqrt{(5-4)^2+(6-3)^2}$ $=\sqrt{1+9}$ $=\sqrt{10}\ \text{units}$ Diagonal $\text{BD}=\sqrt{(3-6)^2+(5-4)^2}$ $=\sqrt{9+1}$ $=\sqrt{10}\ \text{units}$ $\therefore$ AB = BC = CD = AD and diagonal AC = diagonal BD. Thus, ABCD is a quadrilateral in which all sides are equal and the diagonal are equal. Hence, ABCD is a square.
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