Question
The distance between the points $\Big(-\frac{8}{5},2\Big)$ and $\Big(\frac{2}{5},2\Big)$is _________.
✓
Answer
The distance between the points $\Big(-\frac{8}{5},2\Big)$ and $\Big(\frac{2}{5},2\Big)$is 2 unit.
Solution: $(\text{x}_1, \text{y}_1)=\Big(-\frac{8}{5},2\Big)$ $ (\text{x}_2, \text{y}_2)=\Big(\frac{2}{5},2\Big)$ $\text{AB}=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2}$ $\text{AB}=\sqrt{\Big(\frac{2}{5}+\frac{8}{5}\Big)^2+(2-2)}$ $\text{AB}=\sqrt{\Big(\frac{10}{5}\Big)^2+(0)^2}$ $\text{AB}=\sqrt{(2)^2+(0)^2}$ $\text{AB}=\sqrt{4}$ AB = 2 unit
Solution: $(\text{x}_1, \text{y}_1)=\Big(-\frac{8}{5},2\Big)$ $ (\text{x}_2, \text{y}_2)=\Big(\frac{2}{5},2\Big)$ $\text{AB}=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2}$ $\text{AB}=\sqrt{\Big(\frac{2}{5}+\frac{8}{5}\Big)^2+(2-2)}$ $\text{AB}=\sqrt{\Big(\frac{10}{5}\Big)^2+(0)^2}$ $\text{AB}=\sqrt{(2)^2+(0)^2}$ $\text{AB}=\sqrt{4}$ AB = 2 unit
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