Chapter 3.3, Problem 46PPS

(a)

To determine

To sketch: a line $\mathcal{l}$ that contains points $A,B,A^{\prime }$ and $B^{\prime }$ on a coordinate plane.

Explanation of Solution

Draw an inclined line and mark it $\mathcal{l}$ . Spot four focuses on the line and mark them from left to right A, B, A , and B . Draw a line straight down from B and straight opposite A to shape a correct point and mark the vertex of the edge C . Rehash a similar procedure from B and A to frame a correct edge at a point C.

  

(b)

To determine

To describe: $\overline{AC}$ and $\overline{A^{\prime }{C}^{\prime }}$ , and $\overline{BC}$ and $\overline{B^{\prime }{C}^{\prime }}$ .

Answer to Problem 46PPS

  $\overline{AC}$ and $\overline{A^{\prime }{C}^{\prime }}$ are horizontal, and $\overline{BC}$ and $\overline{B^{\prime }{C}^{\prime }}$ are vertical.

Explanation of Solution

Given information:

Add segments to form right triangles $ABC$ and $A^{\prime },B^{\prime },C^{\prime }$ with right angles at $C$ and $C^{\prime }$ .

Calculation:

Now, consider the figure from art (a).

  

$\overline{AC}$ and $\overline{A^{\prime }{C}^{\prime }}$ are horizontal, and $\overline{BC}$ and $\overline{B^{\prime }{C}^{\prime }}$ are vertical.

(c)

To determine

To find: triangles are $ABC$ and $A^{\prime }{B}^{\prime }{C}^{\prime }$ related, and to find that the imply for the slope between any two distinct points on line $\mathcal{l}$ .

Answer to Problem 46PPS

The slope of line through point $A$ and B .

Explanation of Solution

Triangle ABC and A’B’C’ are similar because line $\mathcal{l}$ is a transversal of parallel segments. $\overline{BC}$ and $\overline{B\text{'}C\text{'}}$ .

So, $\angle ABC\cong \angle A\text{'}B\text{'}C\text{'}$ since, corresponding angles of parallel lines are congruent $\angle C\cong \angle C\text{'}$ because all right angles are congruent.

Therefore by AA similarly, $\Delta ABC\sim \Delta A\text{'}B\text{'}C\text{'}$ . Then ratios formed by corresponding sides of similar triangles are equal. The slope of a line defined by through points A and B, is equal to $\frac{B\text{'}C\text{'}}{A\text{'}C\text{'}}$ .

The slope of line through point $A$ and B .