Chapter 2.4, Problem 52PPSE

Solution

(a)

To find:The two right triangles are congruent.

Yes, the two right triangles are congruent.

Given information:

Suppose lines $l_1$ and $l_2$ intersect at the origin. Also, $l_1$ has slope $\frac{y}{x}\left(x>0,y>0\right)$ and $l_2$ has slope $-\frac{x}{y}$ . Then $l_1$ contains $\left(x,y\right)$ and $l_2$ contains $\left(-y,x\right)$ .

  

Calculation:

The line $l_1$ has slope $\frac{y}{x}$ .

The line $l_2$ has slope $-\frac{x}{y}$ .

Since the line $l_1$ and the line $l_2$ are perpendicular to each other.

These two lines are making two right triangles.

  

The side $OA$ and the side $OP$ are congruent.

The angle $m\angle O$ and the angle $m\angle O$ are congruent and the angle $m\angle B$ and the angle $m\angle P$ are congruent.

According to the Angle-Angle-Side concept,

The right triangles $\Delta ABC$ and $\Delta PQR$ are congruent.

Therefore, the two right triangles $\Delta ABC$ and $\Delta PQR$ are congruent.

(b)

To complete: The equation about the angle measures $a,b,c$ and $d$ .

The value of $a$ is equal to ${45}^{\circ }$ , the value of $c$ is equal to ${45}^{\circ }$ , the value of $a+c$ is equal to ${90}^{\circ }$ , andthe value of $b+d$ is equal to ${90}^{\circ }$ .

Given information:

Suppose lines $l_1$ and $l_2$ intersect at the origin. Also, $l_1$ has slope $\frac{y}{x}\left(x>0,y>0\right)$ and $l_2$ has slope $-\frac{x}{y}$ . Then $l_1$ contains $\left(x,y\right)$ and $l_2$ contains $\left(-y,x\right)$ .

  

Calculation:

The line $l_1$ has slope $\frac{y}{x}$ .

The line $l_2$ has slope $-\frac{x}{y}$ .

The right triangles are below:

  

Since the line $l_1$ and the line $l_2$ are perpendicular to each other, so the angle $m\angle O$ is ${90}^{\circ }$ .

  $\begin{array}{l}m\angle a+m\angle b=m\angle O\\ m\angle a+m\angle b={90}^{\circ }\end{array}$

Since the angle $m\angle a$ and the angle $m\angle b$ are equal, so it can be written as below:

  $\begin{array}{c}m\angle a+m\angle a={90}^{\circ }\\ 2\left(m\angle a\right)={90}^{\circ }\\ m\angle a={45}^{\circ }\end{array}$

So, the angle $m\angle b$ is ${45}^{\circ }$ .

In the right triangle ABC,

  $\begin{array}{c}m\angle a+m\angle c+{90}^{\circ }={180}^{\circ }\\ {45}^{\circ }+m\angle c+{90}^{\circ }={180}^{\circ }\\ {135}^{\circ }+m\angle c={180}^{\circ }\\ m\angle c={45}^{\circ }\end{array}$

In the right triangle PQR,

  $\begin{array}{c}m\angle b+m\angle d+{90}^{\circ }={180}^{\circ }\\ {45}^{\circ }+m\angle d+{90}^{\circ }={180}^{\circ }\\ {135}^{\circ }+m\angle d={180}^{\circ }\\ m\angle d={45}^{\circ }\end{array}$

Now, calculate the value of $a+c$ below:

  $\begin{array}{c}a+c={45}^{\circ }+{45}^{\circ }\\ ={90}^{\circ }\end{array}$

Now, calculate the value of $b+d$ below:

  $\begin{array}{c}b+d={45}^{\circ }+{45}^{\circ }\\ ={90}^{\circ }\end{array}$

(c)

To find: The value of $a+b$ .

The value of $a+b$ is equal to ${90}^{\circ }$ .

Given information:

Suppose lines $l_1$ and $l_2$ intersect at the origin. Also, $l_1$ has slope $\frac{y}{x}\left(x>0,y>0\right)$ and $l_2$ has slope $-\frac{x}{y}$ . Then $l_1$ contains $\left(x,y\right)$ and $l_2$ contains $\left(-y,x\right)$ .

  

Calculation:

The line $l_1$ has slope $\frac{y}{x}$ .

The line $l_2$ has slope $-\frac{x}{y}$ .

Since the line $l_1$ and the line $l_2$ are perpendicular to each other, so the angle $m\angle O$ is ${90}^{\circ }$ .

  $\begin{array}{l}m\angle a+m\angle b=m\angle O\\ m\angle a+m\angle b={90}^{\circ }\end{array}$

(d)

To write: The lines $l_1$ and $l_2$ .

The lines $l_1$ and $l_2$ are perpendicular to each other.

Given information:

Suppose lines $l_1$ and $l_2$ intersect at the origin. Also, $l_1$ has slope $\frac{y}{x}\left(x>0,y>0\right)$ and $l_2$ has slope $-\frac{x}{y}$ . Then $l_1$ contains $\left(x,y\right)$ and $l_2$ contains $\left(-y,x\right)$ .

  

Calculation:

The line $l_1$ has slope $\frac{y}{x}$ .

The line $l_2$ has slope $-\frac{x}{y}$ .

Since the line $l_1$ and the line $l_2$ are perpendicular to each other.