Chapter 8.1, Problem 56PS

Solution

(a)

To calculate : The coordinates of $M_1$ and $M_2$ .

The coordinates of $M_1$ are $\left(\frac{x_1}{2},\frac{y_1}{2}\right)$ and of $M_2$ are $\left(\frac{x_1+x_2}{2},\frac{y_1}{2}\right)$ .

Given information :

A triangular sail is being ordered for sailboat. It is planned to sew a thin decorative strip connecting the midpoints $M_1$ and $M_2$ of two sides of the sail.

  

Formula used:

The coordinates of the midpoint $\left(m\right)$ of the line segment joining $A\left(x_1,y_1\right)$ and $B\left(x_2,y_2\right)$ is given by,

  $m=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Calculation :

In the given figure, $M_1$ is the midpoint of $\left(x_1,y_1\right)$ and $\left(0,0\right)$ .

So, the coordinates of $M_1$ will be,

  $\left(\frac{x_1+0}{2},\frac{y_1+0}{2}\right)=\left(\frac{x_1}{2},\frac{y_1}{2}\right)$

  $M_2$ is the midpoint of $\left(x_1,y_1\right)$ and $\left(x_2,0\right)$ . The coordinates of $M_2$ will be,

  $\left(\frac{x_1+x_2}{2},\frac{y_1+0}{2}\right)=\left(\frac{x_1+x_2}{2},\frac{y_1}{2}\right)$

Therefore, the coordinates of $M_1$ are $\left(\frac{x_1}{2},\frac{y_1}{2}\right)$ and of $M_2$ are $\left(\frac{x_1+x_2}{2},\frac{y_1}{2}\right)$ .

(b)

To calculate : The length of the strip and of the sail’s base.

The length of the strip is $\frac{x_2}{2}$ and the length of the sail’s base is $x_2$ .

Hence, the length of the strip is half of the length of the sail’s base.

Given information :

A triangular sail is being ordered for sailboat. It is planned to sew a thin decorative strip connecting the midpoints $M_1$ and $M_2$ of two sides of the sail.

  

Formula used:

The distance $\left(d\right)$ between two points $A\left(x_1,y_1\right)$ and $B\left(x_2,y_2\right)$ is given by,

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

Calculation :

The coordinates of $M_1$ are $\left(\frac{x_1}{2},\frac{y_1}{2}\right)$ and of $M_2$ are $\left(\frac{x_1+x_2}{2},\frac{y_1}{2}\right)$ .

The length of the strip will be the distance between $M_1\text{ and }{M}_2$ ,

  $\begin{array}{c}{M}_1{M}_2=\sqrt{{\left(\frac{x_1+x_2}{2}-\frac{x_1}{2}\right)}^2+{\left(\frac{y_1}{2}-\frac{y_1}{2}\right)}^2}\\ =\sqrt{{\left(\frac{x_2}{2}\right)}^2}\\ =\frac{x_2}{2}\end{array}$

The length of the sail’s base will be the distance between $\left(0,0\right)\text{ and }\left(x_2,0\right)$ ,

  $\begin{array}{c}\sqrt{{\left(x_2-0\right)}^2+{\left(0-0\right)}^2}=\sqrt{{\left(x_2\right)}^2}\\ =x_2\end{array}$

Therefore, the length of the strip is $\frac{x_2}{2}$ and the length of the sail’s base is $x_2$ . Hence, the length of the strip is half of the length of the sail’s base.

(c)

To explain : The length of the strip and the length of the sail's base depends upon the shape of the triangular sail or not.

The length of the strip and the length of the sail's base depends upon the shape of the triangular sail.

Given information :

A triangular sail is being ordered for sailboat. It is planned to sew a thin decorative strip connecting the midpoints $M_1$ and $M_2$ of two sides of the sail.

  

Explanation :

The length of the strip depends up on the coordinates of the sail as $M_1$ and $M_2$ are the midpoints of the two sides of the sail, so as the coordinates of sail will change $M_1$ and $M_2$ will also change.

Therefore, the length of the strip and the length of the sail's base depends upon the shape of the triangular sail.