Question

Find the distance between the points with the given coordinate.

$\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{4}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{3}\mathbf{)}$

Short answer

The distance between the coordinates $(-3,4)$ and$(-2,-3)$ is $\sqrt{50}$units.

Step-by-step solution

Step 1. State the concept of ‘Distance between two points in 2 dimensional plane’.

Let the two points in 2 dimensional plane be $(x_1,y_1)$and $(x_2,y_2)$.

The distance between the points is given as:

$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}\dots \left(1\right)$

Where $d$is the distance.

Step 2. Calculate the distance.

The two points which is given are $(-3,4)$and $(-2,-3)$.

Substitute $(-3,4)$as $(x_1,y_1)$and $(-2,-3)$as $(x_2,y_2)$in $\left(1\right)$to calculate the distance.

$\begin{array}{l}d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}\\ =\sqrt{{(-2-(-3))}^2+{(-3-4)}^2}\\ =\sqrt{{(-2+3)}^2+{(-7)}^2}\\ =\sqrt{{\left(1\right)}^2+{(-7)}^2}\\ =\sqrt{1+49}\\ =\sqrt{50}\end{array}$

Step 3. State the conclusion.

Therefore, distance between the points$(-3,4)$ and $(-2,-3)$is $\sqrt{50}$units.