Chapter 1.3, Problem 106E

To determine

The values of $d_1$ and $d_2$ in term of $m_1\text{and}{m}_2$ .

Answer to Problem 106E

The distance is, $D=\sqrt{2+m_1^2+m_2^2}$

Explanation of Solution

Given information:

  

Formula used:

The distance between two points $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)$ is given by the distant formula,

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

Calculation:

Let us consider the following figure

  

The distance between two points $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)$ is given by the distant formula,

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

The distance $d_1$ is the distance between the points $\left(0,0\right)\text{and}\left(1,m_1\right)$

It is given by,

  $\begin{array}{l}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ d_1=\sqrt{{\left(1-0\right)}^2+{\left(m_1-0\right)}^2}\\ d_1=\sqrt{1+m_1^2}\end{array}$

The distance $d_2$ is the distance between the points (0, 0) and $\left(1,m_2\right)$

It is given by,

  $\begin{array}{l}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ d_1=\sqrt{{\left(1-0\right)}^2+{\left(m_2-0\right)}^2}\\ d_1=\sqrt{1+m_2^2}\end{array}$

Using Pythagoras theorem, the distance between $\left(1,m_1\right)$ and $\left(1,m_2\right)$ is given by

  $\begin{array}{l}D=\sqrt{d_1^2+d_2^2}\\ d_1=\sqrt{1+m_1^2}\\ d_2=\sqrt{1+m_2^2}\\ D=\sqrt{2+m_1^2+m_2^2}\end{array}$

Hence, the distance is, $D=\sqrt{2+m_1^2+m_2^2}$

Conclusion:

The distance is, $D=\sqrt{2+m_1^2+m_2^2}$