Chapter 14.1, Problem 28PSC

To determine

To describe the locus of the midpoint of the ladder in all possible positions.

Answer to Problem 28PSC

  $x^2+y^2=9$

Explanation of Solution

Given:

The triangle with vertices $(x,y),(0,0)$ and $(3,0)$ .

Calculation:

The locus of the midpoint of a $6$ ft ladder in all possible positions is a quarter of a circle with a radius of $3$ ft.

  

Given that drawing to the left, prove algebraically that $x^2+y^2=9$

  

  $\begin{array}{l}{a}^2+b^2=6^2\\ (x,y)=\left(\frac{a+0}{2},\frac{b+0}{2}\right)=\left(\frac{a}{2},\frac{b}{2}\right)\\ x^2+y^2={\left(\frac{a}{2}\right)}^2+{\left(\frac{b}{2}\right)}^2\\ x^2+y^2=\frac{1}{4}\left(a^2+b^2\right)\\ x^2+y^2=\frac{1}{4}\times 36\\ x^2+y^2=9\end{array}$