Prove that A(4, 3), B(6, 4), C(5, 6), D(3, 5) are the vertices of

Q) Prove that A(4, 3), B(6, 4), C(5, 6), D(3, 5) are the vertices of a square ABCD.

Ans: Let’s plot the points on the graph:

Step 1: Now for a quadrilateral ABCD to be a square, required conditions are:

i) its all four sides should be equal i.e. AB = BC = CD = AD

ii) its diagonals should be equal i.e. AC = BD

Step 2: Let’s calculate the lengths of each of the three sides:

We know that the distance between two points (X₁, Y₁) and (X₂, Y₂) is given by:

S = √ (X₂ – X₁)² + (Y₂ – Y₁)²)

∴ AB = $\sqrt{(6 - 4)^2 + (4 - 3)^2} = \sqrt{(4 + 1)}$ = √5

BC = $\sqrt{(5 - 6)^2 + (6 - 4)^2} = \sqrt{(1 + 4)}$ = √5

CD = $\sqrt{(3 - 5)^2 + (5 - 6)^2} = \sqrt{(4 + 1)}$ = √5

and AD = $\sqrt{(3 - 4)^2 + (5 - 3)^2} = \sqrt{(1 + 4)}$ = √5

Since, AB = AC = BC = AD, hence our 1st condition is verified.

Step 3: Let’s check for diagonals now:

∴ AC = $\sqrt{(5 - 4)^2 + (6 - 3)^2} = \sqrt{(1 + 9)}$ = √10

and BD = $\sqrt{(3 - 6)^2 + (5 - 4)^2} = \sqrt{(9 + 1)}$ = √10

Since, AC = BD, hence our 2nd condition is also verified.

Hence, quadrilateral ABCD is a square.

Therefore, the given 4 points re vertices of a square ABCD.

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