**Q)** **Show that the points (- 2, 3), (8, 3) and (6, 7) are the vertices of a right-angled triangle.**

**Ans: **Let’s plot these points on graph, we get:

**Step 1:** Now for a Δ ABC to be an right angled triangle, required condition is:

AB^{2} = AC^{2} + BC^{2}

**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 = **√** (8 – (- 2))^{2 }+ (3 – (3))^{2} ) = **√**(10^{2} + 0) =** 10 units**

Similarly, BC = √ ((8 – 6)^{2 }+ (3 – 7)^{2} ) = √ (4 + 16) = **√ 20 units**

Similarly, AC = √ (6 – (- 2))^{2 }+ (7 – 3)^{2} ) = √ (64 + 16) = **√80** **units**

**Step 3:** Since, (10)^{ 2 }= (√20)^{ 2 } + (√80)^{ 2}

Or AC^{2 }= AB^{2 }+ BC^{2 }

**Therefore, triangle ABC is a right-angled triangle.**

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