Average distance on a triangle 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 .

Verified step by step guidance

1. First, identify the hypotenuse line connecting the points (0, b) and (a, 0). Find its equation by calculating the slope and using point-slope form. The slope \( m \) is given by \( m = \frac{0 - b}{a - 0} = -\frac{b}{a} \). Using point (0, b), the line equation is \( y = -\frac{b}{a} x + b \).
2. Next, consider a point on the x-axis, which has coordinates \( (x, 0) \) where \( x \) ranges from 0 to \( a \). We want to find the vertical distance from this point to the hypotenuse. Since the point is on the x-axis, the vertical distance to the hypotenuse is simply the y-value of the hypotenuse at \( x \), which is \( y = -\frac{b}{a} x + b \).
3. To find the average vertical distance from all points on the x-axis between 0 and \( a \) to the hypotenuse, set up the integral of the vertical distance function over the interval \( [0, a] \) and divide by the length of the interval \( a \). This gives the average distance \( D_{avg} = \frac{1}{a} \int_0^a \left(-\frac{b}{a} x + b\right) dx \).
4. Evaluate the integral \( \int_0^a \left(-\frac{b}{a} x + b\right) dx \) by integrating term-by-term: \( \int_0^a -\frac{b}{a} x \, dx + \int_0^a b \, dx \). Compute each integral separately.
5. After integrating, simplify the expression and divide by \( a \) to find the average vertical distance. You should find that the average distance simplifies to \( \frac{b}{2} \), which is independent of \( a \), as required.

Key Concepts

Equation of a Line (Hypotenuse)

To analyze distances to the hypotenuse, first find its equation. The hypotenuse connects points (0,b) and (a,0), so its slope is -b/a. Using point-slope form, the line equation is y = b - (b/a)x, which describes the hypotenuse for all x between 0 and a.

Vertical Distance from a Point to a Line

The vertical distance from a point on the x-axis (x,0) to the hypotenuse is the difference in y-values since the point lies on y=0. This distance equals the y-coordinate of the hypotenuse at x, which is y = b - (b/a)x. Understanding this helps set up the integral for the average distance.

Average Value of a Function over an Interval

The average vertical distance is the average value of the function y = b - (b/a)x over [0,a]. This is found by integrating the function over the interval and dividing by the interval length a. Calculating this integral shows the average distance equals b/2, independent of a.