Chapter 1: Problem 73
Let \(P=(s, t)\) be a point in the \(x y\) -plane. Let \(P^{\prime}=(t, s)\). Calculate the slope of the line \(\ell^{\prime}\) that passes through \(P\) and \(P^{\prime} .\) Deduce that \(\ell^{\prime}\) is perpendicular to the line \(\ell\) whose equation is \(y=x\). Let \(Q\) be the point of intersection of \(\ell\) and \(\ell^{\prime}\). Show that \(P\) and \(\ell^{\prime}\) are equidistant from \(Q\). (As a result, each of \(P\) and \(P^{\prime}\) is the reflection of the other through the line \(y=x .)\)
Short Answer
Step by step solution
Identify the Points and Line Equation
Calculate the Slope of Line \(\ell^{\prime}\)
Show Perpendicularity of \(\ell^{\prime}\) and \(\ell\)
Find the Intersection Point \(Q\)
Calculate Distance from \(P\) to \(Q\)
Calculate Distance from \(P^{\prime}\) to \(Q\)
Conclude Equidistance and Prove Reflection Property
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Perpendicular Lines
The slope of a line is crucial in determining whether two lines are perpendicular. The product of the slopes of two perpendicular lines is -1. A simple example to remember is if one line has a slope of 1, the perpendicular line will have a slope of -1. This was evident in the problem where line \(\ell\prime\) had a slope of \(-1\) and line \(\ell\), with the equation \(y = x\), had a slope of 1.
- To check for perpendicularity, calculate the slopes of both lines.
- If their product equals -1, the lines are perpendicular.
Distance Formula
In the problem, distances from points \(P=(s,t)\) and \(P'=(t,s)\) to the intersection point \(Q=\left(\frac{s+t}{2}, \frac{s+t}{2}\right)\) were measured. Using the distance formula, it was determined that both points are equidistant from \(Q\).
- This equal distance confirms these points as reflections of each other across the line \(y = x\).
- Revisiting the distance formula can prevent errors in these contexts and ensure accurate calculations.
Slope of a Line
In the exercise, determining the slope of line \(\ell\prime\) involved calculating:\[m_{\ell\prime} = \frac{s - t}{t - s} = -1\]A slope of -1 suggests the line is descending left-to-right, making it perpendicular to upward-sloping lines like \(y = x\), which has a positive slope of 1.
- The slope helps in understanding line orientation and relationships.
- Finding the slope accurately is the first step in analyzing line behaviors.
Intersection Point in Coordinate Geometry
For two lines with equations \(y = mx + b\), equating them gives the intersection x-coordinate:\[ mx_1 + b_1 = mx_2 + b_2\]
Solving this provides coordinates \((x, y)\) of the point where the lines cross. In the problem, intersecting \(y = x\) and line\(\ell\prime = -x + (s + t)\) required solving:\[x = -x + (s + t) \Rightarrow 2x = (s + t) \Rightarrow x = \frac{s + t}{2}\]Thus, the intersection point was \(Q=\left(\frac{s+t}{2}, \frac{s+t}{2}\right)\), pivotal for proving equidistance.
- Simultaneous equations provide a systematic way to find intersections.
- Knowledge of intersection points aids in solving reflection and perpendicularity problems.