Chapter 9.5, Problem 54E

To determine

Plot the point A (1,3) in a coordinate plane. And plot atleast 6 points that are 5 units from A.

Answer to Problem 54E

The graph of the point A (1,3) in a coordinate plane.

  

  $\begin{array}{l}\text{The 6 points 5 units away from the given point A(1,3) are:}\\ \text{(5,0), (-5,0), (5}\text{.58,1), (-5}\text{.58,1), (5}\text{.89,2) and (-5}\text{.89,2)}.\end{array}$

  $$

Explanation of Solution

Given: The point A (1,3) in a coordinate plane.

  

Calculation:

  $\begin{array}{l}Let\text{ the point A(x, 0) is 5 units from the point (1, 3)}\text{.}\\ \text{Use the distance formula:}\\ \text{d = }\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}\\ \Rightarrow 5=\sqrt{{(1-x)}^2+{(3-0)}^2}\Rightarrow 5^2={(}^1+{(}^3\\ \Rightarrow 25={(}^1+9\Rightarrow 16={(}^x\Rightarrow x-1={(}^{16}=\pm 4\\ \Rightarrow x=\pm (1+4)=\pm 5\\ \text{Hence the points are (5,0) and (-5,0)}\\ \text{Similarly, if the point A(x, 1) is 5 units from the point (1, 3)}\text{.}\\ \text{d = }\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}\\ \Rightarrow 5=\sqrt{{(1-x)}^2+{(3-1)}^2}\Rightarrow 5^2={(}^1+4\\ \Rightarrow 21={(}^1\Rightarrow x-1=\sqrt{21}\Rightarrow x-1=\pm 4.58\Rightarrow x=\pm 5.58\\ \Rightarrow \text{Hence the points are (5}\text{.58,1) and (-5}\text{.58,1)}\\ \text{Similarly, if the point A(x, 2) is 5 units from the point (1, 3)}\text{.}\\ \text{d = }\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}\\ \Rightarrow 5=\sqrt{{(1-x)}^2+{(3-2)}^2}\Rightarrow 5^2={(}^1+1\\ \Rightarrow 24={(}^1\Rightarrow x-1=\sqrt{24}\Rightarrow x-1=\pm 4.89\Rightarrow x=\pm 5.89\\ \Rightarrow \text{Hence the points are (5}\text{.89,2) and (-5}\text{.89,2)}\\ \text{Finally the 6 points 5 units away from the given point A(1,3) are:}\\ \text{(5,0), (-5,0), (5}\text{.58,1), (-5}\text{.58,1), (5}\text{.89,2) and (-5}\text{.89,2)}.\end{array}$