Question

cdl, the horizontal distance between points d and l. $|x_{2}-x_{1}| = 15$ blocks, and the vertical distance between points d and l. $|y_{2}-y_{1}| = 19$ blocks.
use the distance formula to determine the distance between points d and l.
$(x_{2}-x_{1})^{2}=$ enter the square of the horizontal distance.
$(y_{2}-y_{1})^{2}=$ enter the square of the vertical distance
$(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}=$ enter the sum of the squares of the horizontal and vertical distances.
$sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}=$ enter the approximate square root of the sum.

Explanation:

Step1: Calculate \((x_{2}-x_{1})^{2}\)

Given \(|x_{2}-x_{1}| = 15\), then \((x_{2}-x_{1})^{2}=15^{2}=225\)

Step2: Calculate \((y_{2}-y_{1})^{2}\)

Given \(|y_{2}-y_{1}| = 19\), then \((y_{2}-y_{1})^{2}=19^{2}=361\)

Step3: Calculate \((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}\)

\((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}=225 + 361=586\)

Step4: Calculate \(\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\)

\(\sqrt{586}\approx24.2\)

Answer:

\((x_{2}-x_{1})^{2}=225\)；\((y_{2}-y_{1})^{2}=361\)；\((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}=586\)；\(\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\approx24.2\)