Chapter 1.3, Problem 63HP

To determine

To find: the value of $y_1$ .

Answer to Problem 63HP

The value of $y_1$ is $y_1=y_2-\sqrt{d^2-{\left(x_2-x_1\right)}^2}$ .

Explanation of Solution

Given information:

Given equation

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

Let us consider the given equation.

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Now, solve the above equation as shown.

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ d^2={\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2\\ d^2-{\left(x_2-x_1\right)}^2={\left(y_2-y_1\right)}^2\\ \sqrt{d^2-{\left(x_2-x_1\right)}^2}=y_2-y_1\text{Square}\text{root}\text{both}\text{sides}\\ \sqrt{d^2-{\left(x_2-x_1\right)}^2}-y_2=-y_1\\ y_1=-\left(\sqrt{d^2-{\left(x_2-x_1\right)}^2}-y_2\right)\\ y_1=y_2-\sqrt{d^2-{\left(x_2-x_1\right)}^2}\end{array}$

Hence, The value of $y_1$ is $y_1=y_2-\sqrt{d^2-{\left(x_2-x_1\right)}^2}$ .