Question

How do you find the distance between point $A\left(-3,5\right)$ and point $B\left(4,-6\right)$ in the coordinate plane?

Answer 1

Find the distance:

The formula for the distance $d$ between the point $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$.

Given points $A\left(-3,5\right)$ and $B\left(4,-6\right)$.

Substitute $x_1=-3,y_1=5,x_2=4,y_2=-6$ in $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ to determine the distance $d$ in between $A$and $B$.

$$\begin{gathered} d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2} \\ \Rightarrow d=\sqrt{{\left(4-\left(-3\right)\right)}^2+{\left(-6-5\right)}^2} \\ \Rightarrow d=\sqrt{{\left(4+3\right)}^2+{\left(-6-5\right)}^2} \\ \Rightarrow d=\sqrt{{\left(7\right)}^2+{\left(-11\right)}^2} \\ \Rightarrow d=\sqrt{49+121} \\ \Rightarrow d=\sqrt{170} \\ \therefore d\approx 13.04\mathrm{unit} \end{gathered}$$

Hence, the distance between point $A\left(-3,5\right)$ and point $B\left(4,-6\right)$ is approximately $13.04\mathrm{unit}$.