Chapter 8.1, Problem 1GP

Solution

To find: what is distance between $\left(3,-3\right)\text{ and }(-1,5)$

Hence, and midpoint formula is given by $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.

Given: two points are $\left(3,-3\right)\text{ and }(-1,5)$

Concept Used:

distance formula is given by

  $\begin{array}{c}\sqrt{{\left(3-(-1)\right)}^2+{\left(-3-5\right)}^2}\\ =\sqrt{{\left(3+1\right)}^2+{\left(-8\right)}^2}\\ =\sqrt{16+64}\\ =\sqrt{80}\\ =8.94427191\\ \approx 8.9\text{ units}\end{array}$

Calculation:

Conclusion:

The distance between $\left(3,-3\right)\text{ and }(-1,5)$ is 8.9 units.