Problem 60 - Calculus Single Variable

Question

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 \(P^{\prime}\) are equidistant from \(Q\). (As a result, \(P\) and \(P^{\prime}\) are reflections of each other through the line \(y=x\)

Step-by-Step Solution

Answer

The slopes of \(\ell'\) and \(\ell\) confirm perpendicularity. \(P\) and \(P'\) are equidistant from \(Q\), showing reflection symmetry over \(y=x\).

Step 1: Understand the Point Coordinates

We have two points in the coordinate plane: \(P = (s, t)\) and \(P' = (t, s)\). These points will help us form the perpendicular slopes and determine the reflection characteristics.

Step 2: Calculate the Slope of Line \( \ell' \)

The slope of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Applying this formula to points \(P = (s, t)\) and \(P' = (t, s)\), we have:\[m = \frac{s - t}{t - s} = \frac{s - t}{-(s - t)} = -1\]

Step 3: Identify Relationship Between Lines \( \ell \) and \( \ell' \)

The line \(\ell\) has an equation \(y = x\), and its slope is 1. Therefore, \(\ell'\) and \(\ell\) are perpendicular because the product of their slopes is \(-1\times 1 = -1\).

Step 4: Find Point of Intersection \(Q\) of \(\ell\) and \(\ell'\)

The line \(\ell\) is \(y = x\). Since \(\ell'\) also passes through the points where \(y = x\), substitute into the point \( (x, x) \) where \(\ell \) and \(\ell'\) intersect.Thus, any point \((x, x)\) satisfies both lines. The specific intersection with the points \((s,t)\) and \((t, s)\) is derived as the point of intersection line, essentially \(Q = \left(\frac{s+t}{2}, \frac{s+t}{2}\right)\), the midpoint since these points equally satisfy both lines.

Step 5: Calculate Distances from \(Q\) to \(P\) and \(P'\)

Distance from \(Q\) to \(P\) can be calculated using the distance formula:\[D_{PQ} = \sqrt{\left(\frac{s+t}{2} - s\right)^2 + \left(\frac{s+t}{2} - t\right)^2}\]Which simplifies to:\[D_{PQ} = \sqrt{\left(\frac{t-s}{2}\right)^2 + \left(\frac{s-t}{2}\right)^2} = \sqrt{\frac{(t-s)^2}{4} + \frac{(s-t)^2}{4}} = \sqrt{\frac{2(t-s)^2}{4}} = \frac{|t-s|}{\sqrt{2}}\]Similarly, applying the formula finds \(D_{P'Q} = \frac{|t-s|}{\sqrt{2}}\), concluding \(P\) and \(P'\) are equidistant from \(Q\).

Key Concepts

Slope of a Line

The slope of a line is a measure of its steepness. Imagine a path that zigzags up a hill—this is similar to how a slope defines a line’s inclination in geometry. Mathematically, the slope between two points

• Given points:
  • Point 1: \((x_1, y_1)\)
  • Point 2: \((x_2, y_2)\)
• Slope formula: \(m = \frac{y_2 - y_1}{x_2 - x_1}\)

This formula calculates how much the line rises vertically compared to how much it runs horizontally. In this exercise, we apply this formula to points \(P = (s, t)\) and \(P' = (t, s)\) to find the slope of line \(\ell'\). The solution showed that:

• Slope of \(\ell':\) \(m = \frac{s-t}{t-s} = -1\)

It means the line from \(P\) to \(P'\) leans in the opposite direction to the line \(y = x\). This difference indicates perpendicularity when the slope product equals \(-1\).

Perpendicular Lines

Perpendicular lines intersect at right angles (90°). When the slopes of two lines multiply to \(-1\), they are perpendicular.
Consider line \(\ell\) with equation \(y = x\), having a slope of \(1\). For a line to be perpendicular to \(\ell\), its slope must be \(-1\).

• Line \(\ell\): Slope = 1
• Line \(\ell'\): Slope = \(-1\)
• Product: \(1 \times -1 = -1\)

When lines \(\ell'\) and \(\ell\) are perpendicular, they create a symmetry in their intersection point. Visualize these lines crossing at a sharp corner, demonstrating the essential geometric feature of perpendicular lines intersecting neatly.

Distance Formula

To find the distance between two points in a plane, we can use the distance formula. It helps us determine how far apart two points are.\[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]In this exercise:

• Find the distance between \(Q\) and \(P\), and \(Q\) and \(P'\).
• With the midpoint of intersection, \(Q = \left(\frac{s+t}{2}, \frac{s+t}{2}\right)\).
• Calculate:
  • \(D_{PQ} = \sqrt{\left(\frac{t-s}{2}\right)^2 + \left(\frac{s-t}{2}\right)^2}\)
  • Distance simplifies to \(\frac{|t-s|}{\sqrt{2}}\).

Both distances \(D_{PQ}\) and \(D_{P'Q}\) are identical, confirming that \(P\) and \(P'\) are equidistant from \(Q\). This equidistance is crucial for understanding reflections in geometry.

Reflection Across a Line

Reflection in geometry acts much like a mirror image, where each point in the original shape has a corresponding point equidistant on the other side of the mirror line. In terms of analytical geometry, reflections help us understand symmetry.
In the problem, \(P\) and \(P'\) being reflections means they are "flipped" over the line \(y = x\).

• Reflection line: \(y = x\)
• For points \((s, t)\) and \((t, s)\), flipping over \(y = x\) means interchanging their coordinates.
• Equidistant: Both \(P\) and \(P'\) have the same distance from the line.

Reflection ensures that geometrically, every feature on one side has an identical counterpart. In this exercise, it is confirmed mathematically through equidistance and coordinate exchanges, illustrating the flawless symmetry of reflections.