Problem 47 - Calculus Single Variable

Question

Find a point that is equidistant from the three points \((3,4),(6,3),\) and (-1,-4)

Step-by-Step Solution

Answer

The point equidistant from all three is (3, -1).

Step 1: Identify the Problem

We need to find a point \(x, y\) that is equidistant from the three given points: \(3, 4\), \(6, 3\), and \(-1, -4\). This point is known as the circumcenter of the triangle formed by these three points.

Step 2: Find the Midpoints

Calculate the midpoints of any two sides of the triangle. Suppose we choose sides \((3, 4)\) to \(6, 3)\) and \((6, 3)\) to \((-1, -4)\):\[\text{Midpoint of } (3, 4) \text{ and } (6, 3): \left(\frac{3+6}{2}, \frac{4+3}{2}\right) = (4.5, 3.5)\]\[\text{Midpoint of } (6, 3) \text{ and } (-1, -4): \left(\frac{6+(-1)}{2}, \frac{3+(-4)}{2}\right) = (2.5, -0.5)\]

Step 3: Calculate the Slopes of the Sides

Find the slopes of the two sides of the triangle: \[\text{Slope of } (3, 4) \text{ to } (6, 3): \frac{3 - 4}{6 - 3} = -\frac{1}{3}\]\[\text{Slope of } (6, 3) \text{ to } (-1, -4): \frac{-4 - 3}{-1 - 6} = 1\]

Step 4: Determine the Perpendicular Bisectors

The slopes of the perpendicular bisectors will be the negative reciprocals of the slopes found in Step 3. Thus:\[\text{Perpendicular slope of } (3, 4) \text{ to } (6, 3): 3\]\[\text{Perpendicular slope of } (6, 3) \text{ to } (-1, -4): -1\]

Step 5: Find the Equations of the Perpendicular Bisectors

Using the point-slope form \(y - y_1 = m(x - x_1)\), we write the equations:\[y - 3.5 = 3(x - 4.5) \Rightarrow y = 3x - 10\]\[y + 0.5 = -1(x - 2.5) \Rightarrow y = -x + 2\]

Step 6: Solve the System of Equations

To find the intersection point of the perpendicular bisectors, solve:\[3x - 10 = -x + 2\]Add \x\ to both sides:\[4x = 12\]Solve for \x\:\[x = 3\]Substitute \x = 3\ into \(y = 3x - 10\):\[y = 3(3) - 10 = -1\]

Step 7: Verify the Solution

The intersection point \(3, -1\) should be equidistant from all three points. Calculate the distance between \(3, -1\) and each point to verify this.

Key Concepts

Equidistant Point

The term "equidistant point" refers to a point that is equally distant from all given points. In the context of a triangle, this specific point is known as the circumcenter. The circumcenter is the point from which the distances to each vertex of the triangle are the same. This means that if you measure the distance from the circumcenter to any vertex, these values will be identical. This property is fundamental when solving problems that require finding a point equidistant from multiple other points, as seen in the exercise. The circumcenter can also be located inside or outside the triangle depending on the nature of the triangle (acute, obtuse, or right). This equidistance is why it acts as the center of a circle, known as the circumcircle, that evenly passes through all vertices of the triangle.
Understanding the concept of equidistant points helps in realizing how geometry uses points and lines to describe balance and symmetry.

Midpoint Calculation

Calculating the midpoint between two points is an essential skill in geometry, particularly in creating segments of equal length. The midpoint is determined by the formula \[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \], where \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the endpoints of a segment. This formula provides the point that is directly in the middle of the two endpoints.
For example, in our exercise, finding midpoints like (4.5, 3.5) and (2.5, -0.5) helps in setting up the perpendicular bisectors, which are crucial for locating the circumcenter. The calculation is straightforward: add the coordinates of the start and end points separately for both axes, then divide each sum by 2.

• Add the x-coordinates and divide by 2 to find the x-coordinate of the midpoint.
• Add the y-coordinates and divide by 2 to find the y-coordinate of the midpoint.

By understanding midpoint calculations, students can place the foundation for constructing segments accurately and delving deeper into topics like bisectors.

Perpendicular Bisectors

In geometry, a perpendicular bisector is a line that divides another line segment into two equal parts at a 90-degree angle. Discovering these bisectors is key for pinpointing the circumcenter. The perpendicular bisector of a segment can be found by calculating the midpoint of the segment and then using the negative reciprocal of the segment's slope to ensure perpendicularity.
Perpendicularity means that if the original segment has a slope \(m\), the bisector will have a slope of \(-\frac{1}{m}\). For instance, if the segment from (3, 4) to (6, 3) has a slope of \(-\frac{1}{3}\), the slope of its perpendicular bisector will be 3. This ensures that the bisector crosses the original segment precisely at a right angle.

• Start with the midpoint of your segment.
• Use the negative reciprocal slope for the bisector.
• Using the point-slope equation, construct the bisector.

This process of creating perpendicular bisectors forms the basis of calculating the intersections needed to locate the circumcenter.

System of Equations

A system of equations is a collection of two or more equations with a common set of unknowns. When dealing with circumcenters, solving a system of equations helps us find the intersection point of the perpendicular bisectors. This intersection provides us with the circumcenter. Given two perpendicular bisectors, we translate them into mathematical equations and then solve these simultaneously.
In our exercise, the equations \(y = 3x - 10\) and \(y = -x + 2\) represent two perpendicular bisectors. By setting them equal and solving for \(x\), we can determine its value in the x-coordinate of the circumcenter. We solved: \[ 3x - 10 = -x + 2 \] to find: \[ x = 3 \].
After finding \(x\), substituting back into either equation helps find \(y\), providing the full solution for the circumcenter.

• Combine equations by aligning a variable.
• Solve for that variable, applying algebraic methods.
• Substitute to find remaining variables' values.

Learning how to solve systems of equations aids students in tackling many real-world and theoretical mathematical problems.