Chapter 1, Problem 1CT

In Problems $1-3$, use $P_1=\left(-1,3\right)$ and $P_2=\left(5,-1\right)$.

Find the distance from $P_1$ to $P_2$.

To determine

To calculate: The distance from $P_1$ to $P_2$, where $P_1=\left(-1,3\right)$ and $P_2=\left(5,-1\right)$.

Answer to Problem 1CT

Solution:

The distance from $P_1$ to $P_2$ is $2\sqrt{13}$.

Explanation of Solution

Given Information:

Points $P_1=\left(-1,3\right)$ and $P_2=\left(5,-1\right)$.

Formula used:

The distance formula $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$.

Calculation:

By comparing $P_1=\left(-1,3\right)$ with $\left(x_1,y_1\right)$ and $P_2=\left(5,-1\right)$ with $\left(x_2,y_2\right)$, it gives

$x_1=-1,y_1=3,x_2=5$ and $y_2=-1$.

By substituting values of $x_1,y_1,x_2$ and $y_2$ in distance formula $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$, it gives

$d=\sqrt{{\left(5-\left(-1\right)\right)}^2+{\left(-1-3\right)}^2}$.

By simplifying,

$\Rightarrow d=\sqrt{{\left(6\right)}^2+{\left(4\right)}^2}$,

$\Rightarrow d=\sqrt{36+16}$,

$\Rightarrow d=\sqrt{52}$,

$\Rightarrow d=2\sqrt{13}$.

Therefore, the distance from $P_1$ to $P_2$ is $2\sqrt{13}$.