Chapter 15, Problem 49CR

To determine

To find: The distance between the given point to the mid-point of the segment.

Answer to Problem 49CR

  $\sqrt{109}$

Explanation of Solution

Given:

  $\begin{array}{l}A=\left(-3,3\right)\text{ is rotated }{90}^o\text{ clockwise}\text{.}\\ C=\left(-2,8\right)\\ D=\left(16,4\right)\end{array}$

Formula Used:

  $\begin{array}{l}\text{Distance}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ \text{Mid-Point}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\end{array}$

Calculation:

  $\begin{array}{l}let,\\ A=\left(-3,3\right)\text{ is rotated }{90}^o\text{ clockwise}\text{.}\\ C=\left(-2,8\right)\\ D=\left(16,4\right)\\ \text{Distance}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ \text{Mid-Point}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ \text{Mid-Point of}\overline{CD}=\left(\frac{-2+16}{2},\frac{8+4}{2}\right)=\left(7,6\right)\\ \text{Distance between Mid-Point of }\overline{CD}\text{ and }A=\sqrt{{\left(7-\left(-3\right)\right)}^2+{\left(6-3\right)}^2}\\ =\sqrt{100+9}\\ =\sqrt{109}\end{array}$

Conclusion:

Thus the required distance is $\sqrt{109}$ .