CBSE Class 10 Mathematics Basic 2023: question 27

Hints

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Step-by-step solution

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1. $AB=\sqrt{(4-1{)}^2+(2-7{)}^2}=\sqrt{9+25}=\sqrt{34}$; $BC=\sqrt{(-1-4{)}^2+(-1-2{)}^2}=\sqrt{25+9}=\sqrt{34}$

   1 mark

2. $CD=\sqrt{(-4+1{)}^2+(4+1{)}^2}=\sqrt{9+25}=\sqrt{34}$; $DA=\sqrt{(1+4{)}^2+(7-4{)}^2}=\sqrt{25+9}=\sqrt{34}$. So $AB=BC=CD=DA$.

   1 mark

3. Diagonals: $AC=\sqrt{(-1-1{)}^2+(-1-7{)}^2}=\sqrt{4+64}=\sqrt{68}$; $BD=\sqrt{(-4-4{)}^2+(4-2{)}^2}=\sqrt{64+4}=\sqrt{68}$. So $AC=BD$. All sides equal and diagonals equal, hence ABCD is a square.

   1 mark

Answer: AB = BC = CD = DA = √34 and AC = BD = √68, so ABCD is a square

Revise the chapter

• Chapter: Coordinate Geometry (notes, formulas, where marks are lost and every past question)
• Topic: Distance Formula