Chapter 10.2, Problem 1AYU

The formula for the distance $d$ from $P_1\text{ = }\left(x_1,y_1\right)$ to $P_2\text{ = }\left(x_2,y_2\right)$ is $d\text{ =}$_______.(p.4)

To determine

To find: The formula for distance $\text{d}$ from $P_1=(x_{1,}{y}_1)$ to $P_2=(x_{2,}{y}_2)$.

Answer to Problem 1AYU

$d\left(P_1,P_2\right)=\sqrt{{\left(x_2-x_{1 }\right)}^2+{\left(y_2-y_1\right)}^2}$

Explanation of Solution

Given:

$P_1=(x_{1,}{y}_1)$ and $P_2=(x_{2,}{y}_2)$

Formula used:

Pythagoras theorem: ${\text{Hypotenuse}}^2={\text{ base}}^2+{\text{ height}}^2$.

Calculation:

Let $(x_{1,}{y}_1)$ denote the coordinate of $P_1$ and $(x_{2,}{y}_2)$ denote the coordinate of $P_2$.

The coordinate of $P_3$ are $(x_{2,}{y}_1)$. The coordinates are shown in the figure below.

The horizontal distance from $P_1$ to $P_3$ is given by absolute value of the difference of $x\text{-coordinate}$ $|x_2-x_1|$. The vertical distance from $P_3$ to $P_2$ is the absolute value of the difference of $y\text{-coordinate}$ $|y_2-y_1|$.

Using Pythagoras theorem, we get,

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

Hence, The distance between two points $P_1=(x_{1,}{y}_1)$ and $P_2=(x_{2,}{y}_2)$, denoted by $d\left(P_1,P_2\right)$.

It is given by $d\left(P_1,P_2\right)=\sqrt{{\left(x_2-x_{1 }\right)}^2+{\left(y_2-y_1\right)}^2}$.