Chapter 12, Problem 1PCT

Find the exact distance d(P₁, P₂) between points P₁ = (−6, 4) and P₂ = (3, −2).

To determine

To find: The distance $d\left(P_1,P_2\right)$ between the points $P_1=\left(-6,4\right){\text{ and P}}_2=\left(3,-2\right)$.

Answer to Problem 1PCT

The distance $d\left(P_1,P_2\right)$ between the points $P_1=\left(-6,4\right){\text{ and P}}_2=\left(3,-2\right)$ is $3\sqrt{13}$.

Explanation of Solution

Formula used:

The Distance formula: The distance between two points $P_1\text{ =}\left(x_1,y_1\right)\text{ and }{P}_2=\left(x_2,y_2\right)$, denoted by $d\left(P_1,P_2\right)$ is $d\left(P_1,P_2\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$.

Calculation:

Substitute $\left(x_1,y_1\right)\text{=}\left(-6,4\right)\text{ and }\left(x_2,y_2\right)=\left(3,-2\right)$ in distance formula.

$\begin{array}{c}d\left(P_1,P_2\right)=\sqrt{{\left(3-\left(-6\right)\right)}^2+{\left(-2-4\right)}^2}\\ =\sqrt{{\left(9\right)}^2+{\left(-6\right)}^2}\\ =\sqrt{81+36}\\ =\sqrt{117}\\ =3\sqrt{13}\end{array}$

Therefore, the distance between $P_1=\left(-6,4\right){\text{ and P}}_2=\left(3,-2\right)$ is $3\sqrt{13}$.