Question

find the distance between the points (-7,2) and (6,2).
\square + \square
= \underline{quadquad} + \underline{quadquad}
= \underline{quadquad} units

\square \boldsymbol{\bigtriangledown} + \square \boldsymbol{\bigtriangledown}
= \square \boldsymbol{\bigtriangledown} + \square \boldsymbol{\bigtriangledown}
= \square unit(s)

Explanation:

Step1: Calculate the absolute values of the differences in x - coordinates

The two points are \((-7,2)\) and \((6,2)\). The formula used is \(|x_2 - x_1|\). Here, \(x_1=-7\) and \(x_2 = 6\). So \(|6-(-7)|=|6 + 7|\) and \(|-7 - 6|=|-(6 + 7)|\). In the given structure, we can also think of it as \(|-7|+|6|\) (since the y - coordinates are the same, the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) with \(y_1 = y_2\) is \(|x_2 - x_1|\), and \(|x_2 - x_1|=|x_1|+|x_2|\) when \(x_1\) and \(x_2\) have opposite signs).
\(|-7|+|6|\)

Step2: Evaluate the absolute values

The absolute value of \(-7\) is \(7\), and the absolute value of \(6\) is \(6\).
\(=7 + 6\)

Step3: Sum the results

\(=13\)

Answer:

\(|-7|+|6|=7 + 6=13\) units