Calculating the Area of a Triangle Formed by Lines: A Comprehensive Guide

Understanding and solving geometric problems is fundamental in mathematics. One such problem involves determining the area of a triangle formed by a given line and a pair of straight lines derived from a quadratic equation. This article provides a step-by-step guide to solving such problems, including the formulas and methods used.

Introduction

In this article, we will explore the process of finding the area of a triangle formed by the line x ? y - 1 0 and the pair of straight lines represented by the equation x^2 - 3xy 2y^2 0. We will cover the necessary steps, explain the algebraic manipulations, and provide a detailed solution.

Step 1: Finding the Equations of the Lines

The given quadratic equation is x^2 - 3xy 2y^2 0. To find the lines this equation represents, we can factor it. The equation can be rewritten as:

[x^2 - 3xy 2y^2 (x - y)(x - 2y) 0]

Setting each factor to zero gives us the following lines:

1. x - y 0

2. x - 2y 0

Step 2: Finding the Points of Intersection

Next, we find the points where the line x - y - 1 0 intersects the lines given in Step 1.

Intersection with x 2y

Substitute x 2y into the equation x - y - 1 0:

[2y - y - 1 0 y - 1 0 y -frac{1}{3}]

Then, substituting y -frac{1}{3} back into x 2y gives:

[x 2 left(-frac{1}{3}right) -frac{2}{3}]

So, the point of intersection is left(-frac{2}{3}, -frac{1}{3}right).

Intersection with x y

Substitute x y into the equation x - y - 1 0:

[y - y - 1 0 -1 0]

Since this equation has no solution, the line x y does not intersect x - y - 1 0. Therefore, the other point of intersection is:

[y -frac{1}{2} x -frac{1}{2}]

So, the point of intersection is left(-frac{1}{2}, -frac{1}{2}right).

Step 3: Finding the Area of the Triangle

We now have the three vertices of the triangle: Aleft(-frac{2}{3}, -frac{1}{3}right), Bleft(-frac{1}{2}, -frac{1}{2}right), and C(-1, 0).

To find the area of the triangle, we can use the determinant method:

[text{Area} frac{1}{2} left| x_1(y_2 - y_3) x_2(y_3 - y_1) x_3(y_1 - y_2) right|]

Substituting the coordinates:

[Aleft(-frac{2}{3}, -frac{1}{3}right), Bleft(-frac{1}{2}, -frac{1}{2}right), C(-1, 0)]

We get:

[text{Area} frac{1}{2} left| -frac{2}{3}(-frac{1}{2} - 0) left(-frac{1}{2}right)(0 - left(-frac{1}{3}right)) (-1) left(-frac{1}{3} - (-frac{1}{2}) right) right|]

Simplifying each term:

[-frac{2}{3} left(-frac{1}{2}right) frac{1}{3}] [-frac{1}{2} left(-frac{1}{3}right) frac{1}{6}] [-1 left(-frac{1}{3} frac{1}{2}right) -1 left(frac{1}{6}right) -frac{1}{6}]

Combining:

[text{Area} frac{1}{2} left( frac{1}{3} - frac{1}{6} - frac{1}{6} right) frac{1}{2} left( frac{1}{3} right) frac{1}{6}]

Therefore, the area of the triangle formed by the given lines is:

[boxed{frac{1}{6}}]

By following these steps, we can solve similar geometric problems effectively using algebraic methods and the formula for the area of a triangle.