Problem 52 - Precalculus

Question

You \(\%\) need to recall the following definitions and results from elementary geometry. In a triangle, a line segment drawn from a vertex to the midpoint of the opposite side is called a median. The three medians of a triangle are concurrent; that is, they intersect in a single point. This point of intersection is called the centroid of the triangle. A line segment drawn from a vertex perpendicular to the opposite side is an altitude. The three altitudes of a triangle are concurrent; the point where the altitudes intersect is the orthocenter of the triangle. This exercise illustrates the fact that the altitudes of a triangle are concurrent. Again, we'll be using \(\triangle A B C\) with vertices \(A(-4,0), B(2,0),\) and \(C(0,6) .\) Note that one of the altitudes of this triangle is just the portion of the \(y\) -axis extending from \(y=0\) to \(y=6 ;\) thus, you won't need to graph this altitude; it will already be in the picture. (a) Using paper and pencil, find the equations for the three altitudes. (Actually, you are finding equations for the lines that coincide with the altitude segments.) (b) Use a graphing utility to draw \(\triangle A B C\) along with the three altitude lines that you determined in part (a). Note that the altitudes appear to intersect in a single point. Use the graphing utility to estimate the coordinates of this point. (c) Using simultaneous equations (from intermediate algebra), find the exact coordinates of the orthocenter. Are your estimates in part (b) close to these values?

Step-by-Step Solution

Answer

The orthocenter is at \((0, \frac{4}{3})\).

Step 1: Identify Altitude Lines

To find the equation of each altitude for triangle ABC, we start by identifying the equations for two altitudes, since the third altitude lies along the y-axis. The first altitude is from vertex A to segment BC, the second is from vertex B to segment AC, and the third is already aligned along the y-axis.

Step 2: Find Slope of BC

We need the slope of the line BC. Using the formula for the slope between two points \((x_1, y_1)\) and \((x_2, y_2)\), it is \(m = \frac{y_2 - y_1}{x_2 - x_1}\). For B(2,0) and C(0,6), the slope is \(m = \frac{6 - 0}{0 - 2} = -3\).

Step 3: Equation of Altitude from A

Since the altitude from A is perpendicular to BC, its slope is the negative reciprocal of BC's slope, which is \[\frac{1}{3}\]. The equation of the line in point-slope form is \(y - y_1 = m(x - x_1)\). Using point A(-4,0), the equation becomes \((y - 0 = \frac{1}{3}(x + 4))\) simplifying to \(y = \frac{1}{3}x + \frac{4}{3}\).

Step 4: Slope of AC

Next, we find the slope of line AC using coordinates A(-4,0) and C(0,6). The slope is \(m = \frac{6 - 0}{0 - (-4)} = \frac{3}{2}\).

Step 5: Equation of Altitude from B

The altitude from B is perpendicular to AC, with a slope \(-\frac{2}{3}\). Using B(2,0), the equation is \(y - 0 = -\frac{2}{3}(x - 2)\), which simplifies to \(y = -\frac{2}{3}x + \frac{4}{3}\).

Step 6: Graph the Triangle and Altitudes

Graph triangle ABC and the two altitudes y = \(\frac{1}{3}x + \frac{4}{3}\) and y = -\(\frac{2}{3}x + \frac{4}{3}\), along with the line x = 0 (the third altitude). The intersection of these lines should appear as a single point on the graph, indicative of the orthocenter.

Step 7: Solve Simultaneous Equations

To find the exact orthocenter, solve the simultaneous equations \(y = \frac{1}{3}x + \frac{4}{3}\) and \(y = -\frac{2}{3}x + \frac{4}{3}\). Setting them equal gives \(\frac{1}{3}x + \frac{4}{3} = -\frac{2}{3}x + \frac{4}{3}\). Simplifying results in \(x = 0\). Substituting \(x = 0\) into any altitude equation, such as \(y = \frac{1}{3}x + \frac{4}{3}\), results in \(y = \frac{4}{3}\). Therefore, the orthocenter is \((0, \frac{4}{3})\).

Step 8: Compare Estimate with Exact Value

Compare the graphically estimated orthocenter with the calculated value of \((0, \frac{4}{3})\). They should be very close or the same if the graphing utility was accurate.

Key Concepts

Medians and Centroids

In any triangle, a median is a special line segment that extends from one of the triangle's vertices to the midpoint of the opposite side. Each triangle has three such medians, and interestingly, they all meet at a single point. This point of intersection is known as the centroid. The centroid serves as the triangle's center of mass or balance point. It divides each median into two segments, with the longer segment being twice the length of the shorter one.

When solving problems involving medians and centroids, it's crucial to identify the midpoint of each side accurately. By finding these midpoints, you can then draw the medians and locate the centroid.

• Median - Line from a vertex to the midpoint of the opposite side.
• Centroid - The intersection point of the medians.

Understanding how medians and centroids work helps in various applications such as finding the balance point of a physical object or calculating the average position of multiple points.

Altitudes and Orthocenters

Altitude lines in a triangle are drawn from each vertex and fall perpendicular to the opposite side or its extension. Therefore, each triangular shape has three altitudes. Where these three altitudes intersect is called the orthocenter. This point can be found inside the triangle for acute triangles, on the hypotenuse for right triangles, and outside for obtuse triangles.

Unique to orthocenters is the fact that they need not be neatly contained within the bounds of a typical triangle. Graphical demonstrations often help visualize altitudes and the orthocenter more effectively.

• Altitude - Line from a vertex perpendicular to the opposite side.
• Orthocenter - The common intersection of the altitudes.

The orthocenter of a triangle can be a challenging concept, as it might not always reside within the triangle's area, contrasting vividly with centroids.

Coordinate Geometry

Coordinate geometry is a method for studying and analyzing geometric figures using a coordinate plane. It allows geometric problems to be solved algebraically, using coordinates and equations. In triangles, this involves assigning coordinates to vertices and making use of various geometric properties.

By utilizing formulas for slopes, distances, and midpoints, one can easily perform operations like calculating altitudes or medians. Coordinate geometry serves as a bridge to link algebra and geometry, making complex problems manageable by systematic processes.

• Coordinate Plane - Tool to study geometry using algebra.
• Geometric Properties - Directions to apply algebraic techniques effectively.

Embracing coordinate geometry in triangle problems offers an analytical edge, enabling clearer calculations and the ability to visualize geometric figures with precision.

Simultaneous Equations

Solving simultaneous equations is a key technique for determining points of intersection between lines in coordinate geometry. To find where two altitudes intersect (an orthocenter), their equations are solved together. If you have equations like \(y = \frac{1}{3}x + \frac{4}{3}\) and \(y = -\frac{2}{3}x + \frac{4}{3}\), setting them equal finds common coordinates.

When two linear equations are equated, they can be simplified to determine specific values for \(x\) and \(y\). This systematic approach allows you to pinpoint exact locations, which is particularly beneficial in geometry.

• Simultaneous Equations - Method to find intersection points.

Using simultaneous equations provides a precise algebraic pathway to solving geometric configurations, elevating understanding from estimations to exact computations.