Chapter 4.3, Problem 31E

To determine

To check whether the paths of puck forms a right triangle.

Answer to Problem 31E

Paths of puck not form a right angled triangle. The claim made by the friend is not correct.

Explanation of Solution

Given information: A hockey puck leaves the blade of a hockey stick, bounce off a wall and travel in a new direction.

Calculation: As hockey puck starts from the point $\mathrm{A}(-4,0)$ bounces on the wall at point $\mathrm{B}(0,8)$ and travel in the direction of the point $\mathrm{C}(-4,16)$ .

slope of AB is:

  $\begin{array}{l}m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{8-0}{0+4}=\frac{8}{4}=2\end{array}$

slope of BC is:

  $m=\frac{16-8}{-4-0}=\frac{8}{-4}=-2$

slope of CA is:

  $m=\frac{16-0}{-4+4}=\frac{16}{\infty }=\infty$

Because not the product of any two slope is $-1$ .

Therefore they are not forming a right angled triangle.

Thus our friend is not correct.

Slope intercept form is $y=mx+c$ .

Put $x=7,y=2\mathrm{and}m=-2$

  $\begin{array}{l}2=(-2)7+c\\ 2=-14+c\\ 2+14=c\\ 16=c\end{array}$

Equation of the line in slope intercept form is:

  $y=-2x+16$

Thus, the equation of the line in slope intercept form is $y=-2x+16$ .