Question

Find the distance $d\left(P_1,P_2\right)$ between the two points $P_1$ and $P_2$. $P_1=\left(-4,-3\right)$; $P_2=\left(2,2\right)$.

Answer 1

Find the distance between two points:

We know that, distance between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by, $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$.

Here, $\left(x_1,y_1\right)=\left(-4,-3\right)$ and $\left(x_2,y_2\right)=\left(2,2\right)$.

Substitute $x_1=-4,x_2=2,y_1=-3$ and $y_2=2$ in $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$.

$\begin{array}{rcl}d& =& \sqrt{{\left(2-\left(-4\right)\right)}^2+{\left(2-\left(-3\right)\right)}^2}\\ & =& \sqrt{{\left(2+4\right)}^2+{\left(2+3\right)}^2}\\ & =& \sqrt{{\left(6\right)}^2+{\left(5\right)}^2}\\ & =& \sqrt{36+25}\\ & =& \sqrt{61}\\ \therefore d& =& 7.81\mathrm{unit}\end{array}$

Hence, distance between two points $P_1=\left(-4,-3\right)$ and $P_2=\left(2,2\right)$ is $d\left(P_1,P_2\right)=7.81\mathrm{unit}$.