Question

Consider the right triangle with vertices \((0,0),(0, b),\) and \((a, 0),\) where \(a > 0\) and \(b > 0 .\) Show that the average vertical distance from points on the \(x\) -axis to the hypotenuse is \(b / 2,\) for all \(a > 0\).

Short answer

Question: Prove that the average vertical distance from points on the x-axis to the hypotenuse of a right-angled triangle with vertices (0,0), (0, b), and (a, 0) is b/2, regardless of a > 0.

Step-by-step solution

Find the equation of the hypotenuse

The triangle has vertices \((0,0),(0, b),\) and \((a, 0),\) so the hypotenuse is the line segment between points \((0, b)\) and \((a, 0).\) Using the two-point form of the equation of a line, the equation of the hypotenuse is: \(y - b = \frac{b - 0}{0 - a}(x - 0)\) Simplifying this expression, we have: \(y = \frac{b}{a}x + b\) This is the equation of the hypotenuse.

Find the vertical distance between a point on the x-axis and the hypotenuse

Let \(P(x, 0)\) be a point on the x-axis, and \(Q(x, y)\) be the corresponding point on the hypotenuse such that the segment \(PQ\) is vertical. Therefore, the coordinate of point \(Q\) is \((x, \frac{b}{a}x + b).\) Since the vertical distance between two points with the same \(x\)-coordinate is the difference in their \(y\)-coordinates, the vertical distance between points \(P\) and \(Q\) is: \(d = \left(\frac{b}{a}x + b - 0\right) = \frac{b}{a}x + b\)

Calculate the average vertical distance

The average vertical distance can be calculated by integrating the vertical distance function over the interval [0, a] and dividing by the length of the interval, which is a: \(\text{Average vertical distance} = \frac{1}{a}\int_0^a \left(\frac{b}{a}x + b\right) dx\) Perform the integration: \(\frac{1}{a}\left(\frac{b}{2a}x^2 + bx\right)\bigg|_0^a = \frac{1}{a}\left(\frac{b}{2a}a^2 + ba\right) = \frac{ab}{2a} + b = \frac{b}{2} \) So the average vertical distance between points on the x-axis and the hypotenuse is \(b / 2,\) for all \(a > 0\).

Key concepts

Right Triangle

A right triangle is a type of triangle that has one angle measuring 90 degrees. This right angle creates the perfect setting for forming various geometric relationships. In the given exercise, we see a right triangle with vertices \((0, 0), (0, b), \) and \((a, 0)\). These points form the right-angle triangle: the vertical side runs along the line from \((0,0)\) to \((0, b)\), the horizontal side from \((0,0)\) to \((a,0)\), and the hypotenuse is the line connecting \((0, b)\) to \((a, 0)\). This layout is crucial because the hypotenuse includes all the slanted lines between two other points and forms the basis for the rest of the problem calculations. The knowledge of how this shape works helps with finding lines, distances, and areas.

Equation of Hypotenuse

To find the equation of the hypotenuse in a right triangle, we rely on the two-point form of a line equation. For points \((0, b)\) and \((a, 0)\), the form becomes:

• The slope of the line (change in y over change in x) becomes \( \frac{b-0}{0-a} = -\frac{b}{a} \).
• Using the line equation \(y - y_1 = m(x - x_1)\), we plug in the known values: \(y - b = -\frac{b}{a}(x - 0)\).

Simplifying, we get \(y = -\frac{b}{a}x + b\). This is the equation of the hypotenuse line, showing the relationship between \(x\) and \(y\) on that line. Knowing this equation is key to finding distances and performing integrals involving the triangle.

Integration

Integration is a mathematical process that helps calculate the area under a curve over a specific interval. In the problem, we need to integrate to find the average vertical distance from the x-axis to the hypotenuse. This involves the following steps:

• Use the vertical distance function \(d = \frac{b}{a}x + b\).
• Integrate over the interval from \(0\) to \(a\): \(\int_0^a (\frac{b}{a}x + b) dx\).

The integration helps accumulate all the small distances into one value, and dividing by \(a\) gives us the average. This powerful tool takes a seemingly difficult task and distills it to one essential result. The outcome \(\frac{b}{2}\) shows the average distance despite varying distances along the way.

Vertical Distance

Vertical distance refers to how far two points are separated in the y-direction. Here, we calculate between a point on the x-axis (\(P(x, 0)\)) and a point on the hypotenuse (\(Q(x, \frac{b}{a}x + b)\)):

• The distance is simply the difference in their y-values: \(d = \frac{b}{a}x + b\).

Using this clear understanding of vertical distance is essential when averaging these distances across the triangle's base. It involves recognizing changes along the x-axis and translating them into the y-direction based on the line's slope and y-intercept. Ultimately, it allows us to evaluate distances from one fundamental perspective, crucial in geometry.