Question
Prove that the points (2, 3), (-4, -6) and $\Big(1, \frac{3}{2}\Big)$ do not form a triangle.
✓
Answer
Let the given points A(2, 3), B(-4, -6) and $\text{C}\Big(1, \frac{3}{2}\Big)$ Then, $\text{AB}=\sqrt{(-4-2)^2+(-6-3)^2}$ $=\sqrt{36+81}$ $=\sqrt{117}=\sqrt{9\times13}$ $=3\sqrt{13 }$ $\text{BC}=\sqrt{(1-(-4))^2+\Big(\frac{3}{2}-(-6)\Big)^2}$ $=\sqrt{25+\Big(\frac{3}{2}+6\Big)^2}$ $=\sqrt{25+\Big(\frac{15}{2}\Big)^2}$ $=\sqrt{25+\frac{225}{4}}$ $=\sqrt{\frac{100+225}{4}}$ $=\sqrt{\frac{335}{4}}$ $=\sqrt{83.75}$ $\text{AC}=\sqrt{(1-2)^2+\Big(\frac{3}{2}-3\Big)^2}$ $=\sqrt{1+\Big(\frac{3-6}{2}\Big)^2}$ $=\sqrt{1+\frac{9}{4}}=\sqrt{\frac{13}{4}}$$=\sqrt{3.25}$
We know that, In triangle sum of any two sides is greater than the third side. Since, $\text{AC} + \text{AB} \ngtr \text{BC}$Here, Sum of two sides is not greater than the third side.
Therefore, Points (2, 3), (-4, -6) and $\Big(1, \frac{3}{2}\Big)$ do not form a triangle.
We know that, In triangle sum of any two sides is greater than the third side. Since, $\text{AC} + \text{AB} \ngtr \text{BC}$Here, Sum of two sides is not greater than the third side.
Therefore, Points (2, 3), (-4, -6) and $\Big(1, \frac{3}{2}\Big)$ do not form a triangle.
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