Question

Prove that the points A(0, 1), B(1,4), C(4,3) and D (3,0) are the vertices of a
square.
​

Answer 1

Answer:-

Given:-

• The points of vertices of square are A(0,1),B (1,4),C(4,3) and D(3,0).

To find:-

• Prove the points

Required formula:-

• Distance between two points = √(x₂ - x₁)² + (y₂ - y₁)²

Here,Finding the distance between the points AB;

$\\ : \implies \sf{AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} } } \\ \\ : \implies \sf{AB = \sqrt{ {(1 - 0)}^{2} + {(4 - 1)}^{2} } } \\ \\ : \implies \sf{AB = { \sqrt{ {(1)}^{2} + {(3)}^{2} } }} \\ \\ : \implies \sf{AB = \sqrt{1 + 9} } \\ \\ : \implies \underline {\boxed{\frak{AB = \sqrt{10}\: units}}} \: \red{ \bigstar}$

Now, finding the distance between the points BC;

$\\ :\implies\sf{BC = \sqrt{ {(4 - 1)}^{2} + {(3 - 4)}^{2}}} \\ \\ :\implies\sf{BC = \sqrt{ {(3)}^{2} + {( - 1)}^{2} } } \\ \\ :\implies\sf{BC = \sqrt{9 + 1}} \\ \\ :\implies \underline{ \boxed{\frak{BC = \sqrt{10} \: units}}} \: \green{ \bigstar}$

finding the distance between the points CD;

$\\ :\implies\sf{CD = \sqrt{{(3-4)}^{2}+{(0-3)}^{2}}} \\ \\ :\implies\sf{CD = \sqrt{{(-1)}^{2}+{(-3)}^{2}}}\\ \\ :\implies\sf{CD = \sqrt{1+9}}\\ \\ :\implies\underline{\boxed{\frak{CD = \sqrt{10}\: units}}} \: \blue{ \bigstar}$

Finally,finding the distance between the points DA;

$\\ :\implies\sf{DA = \sqrt{{(0-3)}^{2}+{(1-0)}^{2}}}\\ \\ :\implies\sf{DA = \sqrt{{(-3)}^{2}+{(1)}^{2}}} \\ \\ :\implies\sf{DA = \sqrt{9 + 1}} \\ \\ :\implies\underline {\boxed{\frak{DA = \sqrt{10} \: units}}} \: \pink{ \bigstar}$

Therefore,

• AB = BC = CD = DA are equal.

• Hence,Proved.