**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 (X_{1}, Y_{1}) and (X_{2}, Y_{2}) is given by:

S = **√ **(X_{2} – X_{1})^{2 } + (Y_{2} – Y_{1})^{2 })

∴ AB = =** √5**

BC = =** √5**

CD = =** √5**

and AD = =** √5**

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

**Step 3:** Let’s check for diagonals now:

∴ AC = =** √10**

and BD = =** √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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