Chapter 10.3, Problem 1AYU

The distance $d$ from $P_1=\left(2,-5\right)$ to $P_2=\left(4,-2\right)$ is $d=$ ______. (p.4)

To determine

To fill: The distance $d$ from $P_1=\left(2,-5\right)$ to $P_2=\left(4,-2\right)$.

Answer to Problem 1AYU

Solution:

The distance $d$ from $P_1=\left(2,-5\right)$ to $P_2=\left(4,-2\right)$.

$d=\sqrt{13}$

Explanation of Solution

Given:

$P_1=\left(2,-5\right)$ and $P_2=\left(4,-2\right)$.

Formula Used:

$d\left(P_1, P_2\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

$d\left(P_1, P_2\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

$d\left(P_1, P_2\right)=\sqrt{{\left(4-2\right)}^2+{\left(-2-\left(-5\right)\right)}^2}$

$d\left(P_1, P_2\right)=\sqrt{{\left(2\right)}^2+{\left(3\right)}^2}$

$d\left(P_1, P_2\right)=\sqrt{4+9}\text{ = }\sqrt{13}$

$d\left(P_1, P_2\right)=\sqrt{13}$