Finding the Equation of a Line Through Given Points Without Using Slopes

In this article, we will explore a method to find the equation of a line passing through given points without resorting to the traditional slope approach. We will use alternative forms such as the point-point form and cross product for 3D geometry. By understanding these methods, you can gain a deeper insight into the duality between points and lines, which is a fascinating aspect of geometry.

Point-Point Form Approach

The equation of a line passing through two points, say (x1, y1) and (x2, y2), can be given in the point-point form:

(frac{y - y1}{x - x1} frac{y2 - y1}{x2 - x1})

Given points: (-1, 1) and (2, -4)

Step-by-Step Solution

1. Substitute the points into the equation:

(frac{y - 1}{x - (-1)} frac{-4 - 1}{2 - (-1)})

2. Simplify the equation:

(frac{y - 1}{x 1} frac{-5}{3})

3. Cross multiply to solve for y:

(3(y - 1) -5(x 1))

4. Expand and rearrange to get the line equation:

(3y - 3 -5x - 5)

(5x 3y -2)

The final equation of the line is: (5x 3y -2)

Cross Product Method for 3D Geometry

The cross product method allows us to determine a line that passes through two points in 3D space. This method is particularly useful in projects involving vector calculations and geometry.

Three Points Not in a Line

Given three points, if they do not lie on the same line, the difference between the two sides (c - ay - b - d - bx - a) is twice the signed area of the resulting triangle.

Example with 3 Points

Given points: (-1, 1), (2, -4), and (3, -2)

1. Set up the cross product for the vectors from the first point to the other two points:

(left| begin{array}{ccc} 2 - (-1) -4 - 1 3 - (-1) -2 - 1 end{array} right| left| begin{array}{ccc} 3 -5 4 -3 end{array} right|)

2. Calculate the determinant:

(det (3)(-3) - (-5)(4) -9 20 11)

3. The resulting vector ((b - c, c - a, a - b)) gives the normal vector of the plane:

((b - c, c - a, a - b) (4 - 11, -4 - 2, 1 - 3) (-7, -6, -2))

The normal vector ((7, 6, 2)) corresponds to the plane equation:

(7x 6y 2z d)

Using one of the points, we find d:

(7(-1) 6(1) 2(-4) d)

(d -7 6 - 8 -9)

The final plane equation is: (7x 6y 2z -9)

Complete Duality Between Points and Lines

The concept of duality in geometry implies that points and lines have a reciprocal relationship. Points can be considered as lines, and lines as points. Using this duality, we can represent ( 2, -1) and ( 3, -4) as lines:

(2x - y 0) and (3x - 4y 0)

Using the cross product, we get:

(left| begin{array}{ccc} 3 -1 -4 -2 end{array} right| (3)(-2) - (-1)(-4) -6 - 4 -10)

The resulting vector ((d, e, f)) gives the equation of the line:

(3x - 11y -10)

Rearranging, we get:

(3x - 11y 10)

Additional Example

Given points: (2, -1) and (3, -4)

1. Calculate the slope (m):

(m frac{-4 - (-1)}{3 - 2} frac{-3}{1} -3)

2. Point-slope form: (y - y1 m(x - x1))

Substitute (2, -1) into the equation:

(y 1 -3(x - 2))

Rearrange to get the line equation:

(y -3x 6 - 1)

(y -3x 5)

The final equation of the line is: (3x y 5)

Conclusion

In conclusion, finding the equation of a line through given points can be approached in multiple ways, each offering unique insights into the structure of geometry. Whether through the point-point form or the cross product method, these techniques provide a robust foundation for understanding the relationship between points and lines. By applying these methods, you can effectively tackle a variety of problems in geometry and 3D vector spaces.