CBSE Class 10 Mathematics Standard 2023: question 23

Hints

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Step-by-step solution

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1. Let the points be $A(-2,3)$, $B(8,3)$ and $C(6,7)$. Use the distance formula:
   $AB=\sqrt{(8+2{)}^2+(3-3{)}^2}=10$
   $BC=\sqrt{(6-8{)}^2+(7-3{)}^2}=\sqrt{4+16}=\sqrt{20}$

   1 mark

2. $AC=\sqrt{(6+2{)}^2+(7-3{)}^2}=\sqrt{64+16}=\sqrt{80}$

   ½ mark

3. $B{C}^2+A{C}^2=20+80=100=A{B}^2$.
   By the converse of Pythagoras theorem, the triangle is right-angled at $C$. Hence the points are the vertices of a right-angled triangle.

   ½ mark

Answer: $A{B}^2=B{C}^2+A{C}^2$ ($100=20+80$), so the triangle is right-angled at C.

Revise the chapter

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