Chapter 13.5, Problem 19WE

(a)

To determine

To find: The coordinate of the midpoint $M$ in right $\Delta OAT$ .

Answer to Problem 19WE

The coordinate of the midpoint $M$ is $\left(-3,4\right)$ .

Explanation of Solution

Given information:

Consider the given $\Delta OAT$ .

  

Calculation:

Apply midpoint formula to calculate the coordinates of point $M$ .

  $\begin{array}{c}M=\left(\frac{-6+0}{2},\frac{0+8}{2}\right)\\ =\left(\frac{-6}{2},\frac{8}{2}\right)\\ =\left(-3,4\right)\end{array}$

Conclusion:

Therefore, the coordinate of the midpoint $M$ is $\left(-3,4\right)$ .

(b)

To determine

To find: The length of the segments $MA$ , $MT$ , and $MO$ in right $\Delta OAT$ . Compare the lengths.

Answer to Problem 19WE

The length of the segments $MA$ , $MT$ , and $MO$ is equal and it is 5 units.

Explanation of Solution

Given information:

Consider the given $\Delta OAT$ .

  

Calculation:

From part (a), the coordinate of the midpoint $M$ is $\left(-3,4\right)$ .

Apply distance formula between point $M$ and $A$ to calculate the length.

  $\begin{array}{c}MA=\sqrt{{\left(0-\left(-3\right)\right)}^2+{\left(8-4\right)}^2}\\ =\sqrt{{\left(3\right)}^2+{\left(4\right)}^2}\\ =\sqrt{25}\\ =5\text{units}\end{array}$

Similarly, apply distance formula between point $M$ and $T$ to calculate the length.

  $\begin{array}{c}MT=\sqrt{{\left(-3-\left(-6\right)\right)}^2+{\left(4-0\right)}^2}\\ =\sqrt{{\left(-3\right)}^2+{\left(4\right)}^2}\\ =\sqrt{25}\\ =5\text{units}\end{array}$

Since, the point $O$ is located at the origin. So, the coordinate of the point $O$ is $\left(0,0\right)$ . Apply distance formula between point $M$ and $O$ to calculate the length.

  $\begin{array}{c}MO=\sqrt{{\left(-3-0\right)}^2+{\left(4-0\right)}^2}\\ =\sqrt{{\left(-3\right)}^2+{\left(4\right)}^2}\\ =\sqrt{25}\\ =5\text{units}\end{array}$

From the above, it is clear that the length of the segments $MA$ , $MT$ , and $MO$ is equal and it is 5 units.

Conclusion:

Therefore, the length of the segments $MA$ , $MT$ , and $MO$ is equal and it is 5 units.

(c)

To determine

To write: The theorem stated in the chapter for the midpoint.

Answer to Problem 19WE

The theorem stated in the chapter is given below.

Explanation of Solution

Calculation:

In $\Delta OAT$ , the point $M$ is the midpoint of the segment $AT$ . According to the theorem, “the midpoint of the hypotenuse of a right triangle is always equidistance from the vertices.”

Conclusion:

Therefore, the theorem stated in the chapter is given above.

(d)

To determine

To find: The equation of the circle that circumscribes $\Delta OAT$ .

Answer to Problem 19WE

The equation of the circle is ${\left(x+3\right)}^2+{\left(y-4\right)}^2=25$ .

Explanation of Solution

Given information:

From part (a), the coordinates of the midpoint is $\left(-3,4\right)$ . From part (b), the length of segment $MO$ is 5 units.

Calculation:

Let the center of the circle is $O$ and its radius is $MO$ .

Write the expression for the standard form of the circle.

  ${\left(x-a\right)}^2+{\left(y-b^2\right)}^2=r^2$

Where,

• $r$ is the radius of the circle.
• $a$ and $b$ are the values of a point that lie on the circle.

Since, the midpoint $M$ lies on the circumference of the circle. So, it satisfied the equation of the circle.

Substitute the values in the above expression.

  $\begin{array}{c}{\left(x-\left(-3\right)\right)}^2+{\left(y-4\right)}^2=5^2\\ {\left(x+3\right)}^2+{\left(y-4\right)}^2=25\end{array}$

Conclusion:

Therefore, the equation of the circle is ${\left(x+3\right)}^2+{\left(y-4\right)}^2=25$ .