Q18. - Algebra 1

Question

Find the distance between the points with the given coordinate.

$\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{2}\mathbf{)}$

Step-by-Step Solution

Answer

The distance between the coordinates $(-1,-1)$ and $(3,2)$ is 5 units.

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 $(-1,-1)$ and $(3,2)$.

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

 $\begin{array}{l}d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}\\ =\sqrt{{(3-(-1))}^2+{(2-(-1))}^2}\\ =\sqrt{{(3+1)}^2+{(2+1)}^2}\\ =\sqrt{{\left(4\right)}^2+{\left(3\right)}^2}\\ =\sqrt{16+9}\\ =\sqrt{25}\\ =5\end{array}$

Step 3. State the conclusion.

Therefore, distance between the points $(-1,-1)$ and $(3,2)$ is 5 units.