Chapter 9.4, Problem 9PSA

a.

To determine

To calculate:The coordinates of C

Answer to Problem 9PSA

The coordinate of C is $\left(2,3\right)$

Explanation of Solution

Given: $\overline{AC}\parallel y-axis$ and $\overline{CB}\parallel x-axis$ are provided in the question which will help to find the solution.

Concept used: Here, we are using Pythagoras theorem.

  ${\left(Hypotenuse\right)}^2={\left(perpendicular\right)}^2+{\left(base\right)}^2$

Calculation: Here, the formula is

  ${\left(Hypotenuse\right)}^2={\left(perpendicular\right)}^2+{\left(base\right)}^2$

Here, hypotenuse = AB

Perpendicular = AC

And, base = CB

${\left(AB\right)}^2={(AC)}^2+{(CB)}^2$

  $64+36={(y-11)}^2+{(3-y)}^2+36$

  $64=y^2+121-22y+9+y^2-6y$

  $2y^2-28y+66=0$

  $y^2-14y+33=0$

  $\left(y-11\right)\left(y-3\right)=0$

  $y=11,3$

Conclusion: Hence, the coordinate of C is $\left(2,3\right)$

b.

To determine

To calculate: The lengths of AC and CB

Answer to Problem 9PSA

The lengths of AC and CB are $8$ and $6$

Explanation of Solution

Given: $\overline{AC}\parallel y-axis$ and $\overline{CB}\parallel x-axis$ are provided in the question which will help to find the solution.

Concept used: Here, we are using distance formula.

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

Calculation: Here, the formula is

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

  $AC=\sqrt{{\left(3-11\right)}^2+{\left(2-2\right)}^2}$

  $AC=8$

And, $CB=\sqrt{{\left(3-3\right)}^2+{\left(8-2\right)}^2}$

  $CB=6$

Conclusion: Hence, the lengths of AC and CB are $8$ and $6$

c.

To determine

To calculate:The length of AB

Answer to Problem 9PSA

The length of AB is $10$

Explanation of Solution

Given: $\overline{AC}\parallel y-axis$ and $\overline{CB}\parallel x-axis$ are provided in the question which will help to find the solution.

Concept used: Here, we are using distance formula.

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

Calculation: Here, the formula is

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

  $AB=\sqrt{{\left(3-11\right)}^2+{\left(8-2\right)}^2}$

  $AB=\sqrt{64+36}$

  $AB=10$

Conclusion: Hence, the length of AB is $10$

d.

To determine

  $AB=\sqrt{{\left(8-2\right)}^2+{\left(11-3\right)}^2}$

Answer to Problem 9PSA

AB is equal to $\sqrt{{\left(8-2\right)}^2+{\left(11-3\right)}^2}$

Explanation of Solution

Given: $\overline{AC}\parallel y-axis$ and $\overline{CB}\parallel x-axis$ are provided in the question which will help to find the solution.

Concept used: Here, we are using distance formula.

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

Calculation: Here, the formula is

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

  $AB=\sqrt{{\left(3-11\right)}^2+{\left(8-2\right)}^2}$

Conclusion: Hence, AB is equal to $\sqrt{{\left(8-2\right)}^2+{\left(11-3\right)}^2}$