Chapter 1.8, Problem 35E

To determine

To calculate: The closer point among $P\left(3,1\right)$ and $Q\left(-1,3\right)$ to the point $R\left(-1,-1\right)$ .

Answer to Problem 35E

Thecloser point tothe point $R\left(-1,-1\right)$ is $Q\left(-1,3\right)$ .

Explanation of Solution

Given information:

The points $P\left(3,1\right)$ and $Q\left(-1,3\right)$ .

Formula used:

Distance $d\left(A,B\right)$ between two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Calculation:

Consider the provided set of points $P\left(3,1\right)$ and $Q\left(-1,3\right)$ . Also the point $R\left(-1,-1\right)$ .

Recall that the distance $d\left(A,B\right)$ between two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Evaluate the distance between $P\left(3,1\right)$ and $R\left(-1,-1\right)$ .

  $\begin{array}{c}d\left(R,P\right)=\sqrt{{\left(-1-3\right)}^2+{\left(-1-1\right)}^2}\\ =\sqrt{16+4}\\ =\sqrt{20}\\ =4.47\end{array}$

Next evaluate the distance between $Q\left(-1,3\right)$ and $R\left(-1,-1\right)$ .

  $\begin{array}{c}d\left(Q,R\right)=\sqrt{{\left(-1-\left(-1\right)\right)}^2+{\left(-1-3\right)}^2}\\ =\sqrt{16}\\ =4\end{array}$

Observe that $d\left(R,P\right)>d\left(Q,P\right)$ .

That is the distance between the point $Q\left(-1,3\right)$ and the $R\left(-1,-1\right)$ is less than the distance the point $P\left(3,1\right)$ and the $R\left(-1,-1\right)$ .

Therefore, the point $Q\left(-1,3\right)$ is near to $R\left(-1,-1\right)$ .

Thus, the closer point to the point $R\left(-1,-1\right)$ is $Q\left(-1,3\right)$ .