Problem 37 - Analytical Geometry: 2D and 3D

Question

Show that the area of the triangle formed by the lines \(y=m_{1} x+c_{1}, y=m_{2} x+\) \(c_{2}\), and \(y=m_{3} x+c_{3}\) is \(\frac{1}{2}\left[\frac{\left(c_{2}-c_{3}\right)^{2}}{m_{2}-m_{3}}+\frac{\left(c_{3}-c_{1}\right)^{3}}{m_{3}-m_{1}}+\frac{\left(c_{1}-c_{2}\right)^{2}}{m_{1}-m_{2}}\right] .\)

Step-by-Step Solution

Answer

Provide a simplified expression for the area. Answer: The area of the triangle formed by the intersection of the given lines is given by the expression: \(\frac{1}{2}\left[\frac{\left(c_{2}-c_{3}\right)^{2}}{m_{2}-m_{3}}+\frac{\left(c_{3}-c_{1}\right)^{3}}{m_{3}-m_{1}}+\frac{\left(c_{1}-c_{2}\right)^{2}}{m_{1}-m_{2}}\right]\).

Step 1: Find Coordinates of Vertices

Let the given lines be \(L_1: y = m_1x+c_1\), \(L_2: y = m_2x+c_2\), and \(L_3: y = m_3x+c_3\). To find the vertices of the triangle, we need to solve for the points of intersections of the lines in pairs. Let's denote the vertices of the triangle as \(A\), \(B\), and \(C\). 1. To find the intersection point of \(L_1\) and \(L_2\), equate the expressions for \(y\): \(m_1x+c_1 = m_2x+c_2\). Solve for \(x\) to find the \(x\)-coordinate of the intersection point \(A\): \(x_A = \frac{c_2-c_1}{m_1-m_2}\), Now substitute this \(x_A\) back into one of the line equations (e.g., \(L_1\)) to find \(y_A\): \(y_A = m_1 x_A + c_1 = m_1\left(\frac{c_2-c_1}{m_1-m_2}\right)+c_1\). Thus, the intersection point A has coordinates \((x_A, y_A) = \left(\frac{c_2-c_1}{m_1-m_2}, m_1\left(\frac{c_2-c_1}{m_1-m_2}\right)+c_1\right)\). 2. Similarly, we will find the intersection point \(B\) of \(L_2\) and \(L_3\), and \(C\) of \(L_1\) and \(L_3\).

Step 2: Calculate the Area Using the Determinant Formula

Use the coordinates of vertices A, B, and C to calculate the area of the triangle. The area of the triangle formed by the points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be found using the determinant formula: Area \(= \frac{1}{2} |(x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2))|\). Plug the coordinates of A, B, and C from Step 1 into the formula and get the expression for the area.

Step 3: Simplify the Expression for the Area

Use algebraic simplification to match the obtained expression for the area to the given target formula. This step will involve careful simplification and manipulation of algebraic terms to arrive at the final formula. \(\frac{1}{2}\left[\frac{\left(c_{2}-c_{3}\right)^{2}}{m_{2}-m_{3}}+\frac{\left(c_{3}-c_{1}\right)^{3}}{m_{3}-m_{1}}+\frac{\left(c_{1}-c_{2}\right)^{2}}{m_{1}-m_{2}}\right]\).

Key Concepts

Coordinates of Vertices

Understanding the coordinates of vertices in analytical geometry is crucial when solving problems related to shapes such as triangles. Coordinates are essentially the position of points on a plane that are determined by numerical values along axes. In the case of a triangle, the coordinates of the vertices are found by locating the intersection points of the lines that form the sides of the triangle.

To do this practically, one takes two lines at a time and solves their equations simultaneously. Since each pair of lines intersects at a single point, this process gives the exact coordinates for each vertex. Generally, the algebra involved here is solving linear equations, and one can find either the x-coordinate or y-coordinate first, then substitute back to find the remaining coordinate.

When solving these problems, watch out for special cases such as parallel lines, which do not meet and therefore do not form a triangle, or coincident lines which lie on top of each other, forming an infinite number of intersection points.

Intersection of Lines

The intersection of lines is a foundational concept in analytical geometry, representing the point at which two lines meet. Each intersection defines a vertex of a polygon, such as a triangle in the case of our exercise. To find the point of intersection, we set the equations of the two lines equal to each other, as each line's equation represents all the points on that line. This can be done for both the x and y coordinates.

When finding the intersection points algebraically, one typically isolates one variable and then substitutes it into the other equation, which simplifies to an expression that gives one of the coordinates of the intersection point. This process is repeated for each pair of lines that form the triangle. A clear understanding of this concept drastically improves one's ability to solve for the vertices in any polygonal shape analysis.

Determinant Formula for Area

The determinant formula for the area of a triangle is an elegant expression that allows one to calculate the area using only the coordinates of the vertices. It is derived from the basic principles of vector algebra and the concept of the cross product

To apply this formula, you need to arrange the vertices' coordinates in a specific order and perform calculations akin to the determinant of a 3x3 matrix. The key steps involve multiplication and subtraction of the coordinates. Importantly, the absolute value is considered to ensure the area is positive. This formula is extremely powerful as it bypasses the need to determine the base and height of the triangle or other more complex geometric measures.

Algebraic Simplification

Algebraic simplification is the process of reducing equations to their simplest form. This often involves combining like terms, factoring, expanding, and using algebraic properties such as distributive, associative, and commutative laws.

In our context, simplification means manipulating the calculated area's expression to match a given target formula. This could involve combining fractions, squaring differences, and careful cancellations. Such algebraic gymnastics require one to be meticulous to prevent errors while simplifying complex algebraic expressions into more digestible forms that reveal the beauty and symmetry inherit in mathematics, as with the final area formulation of the triangle in the exercise.