Question

The distance between the points$\left(a,b\right)\text{and}\left(c,d\right)$ is .So the distance between$\left(1,2\right)\text{and}\left(7,10\right)$ is .

Short answer

The distance between the points$\left(a,b\right)\text{and}\left(c,d\right)$ is $\sqrt{{\left(c-a\right)}^2+{\left(d-b\right)}^2}$. So the distance between$\left(1,2\right)\text{and}\left(7,10\right)$ is 10 units.

Step-by-step solution

Step 1. Apply the concept of the distance formula.

Distance$d\left(A,B\right)$ between two points$A=\left(x_1,y_1\right)\text{and}B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$.

Step 2. Compute the distance between the points.

Consider the provided points $\left(a,b\right)\text{and}\left(c,d\right)$.

Evaluate the distance between the points.

Let $\left(x_1,y_1\right)=\left(a,b\right)\text{and}\left(x_2,y_2\right)=\left(c,d\right)$.

$\begin{array}{c}d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(c-a\right)}^2+{\left(d-b\right)}^2}\end{array}$
Therefore, the distance between the points$\left(a,b\right)\text{and}\left(c,d\right)$ is $\sqrt{{\left(c-a\right)}^2+{\left(d-b\right)}^2}$.

Step 3. Compute the distance between the points. 

Consider the provided points $\left(1,2\right)\text{and}\left(7,10\right)$.

Evaluate the distance between the points.

Let $\left(a,b\right)=\left(1,2\right)\text{and}\left(c,d\right)=\left(7,10\right)$.

$\begin{array}{c}d\left(A,B\right)=\sqrt{{\left(c-a\right)}^2+{\left(d-b\right)}^2}\\ =\sqrt{{\left(7-1\right)}^2+{\left(10-2\right)}^2}\\ =\sqrt{6^2+8^2}\\ =\sqrt{36+64}\end{array}$

Simplify it further as.

$\begin{array}{c}d\left(A,B\right)=\sqrt{36+64}\\ =\sqrt{100}\\ =10\end{array}$

Therefore, the distance between the points $\left(1,2\right)\text{and}\left(7,10\right)$is 10 units.