Problem 55 - Analytical Geometry: 2D and 3D

Question

If the lines \(p_{1} x+q_{1} y=1, p_{2} x+q_{2} y=1\), and \(p_{3} x+q_{3} y=1\) are concurrent, prove that the points \(\left(p_{1}, q_{1}\right),\left(p_{2}, q_{2}\right)\), and \(\left(p_{3}, q_{3}\right)\) are collinear.

Step-by-Step Solution

Answer

Short Answer: Yes, the points (p1, q1), (p2, q2), and (p3, q3) are collinear. This is because the determinant of the points, which is equal to the determinant expressing the concurrency condition, is zero, implying that the points lie on a straight line.

Step 1: Write down the concurrency condition for the three lines

Since the three lines are concurrent, they intersect at one single point, say (x, y). Let's substitute (x, y) into each line equation: 1) \(p_{1}x + q_{1}y = 1\) 2) \(p_{2}x + q_{2}y = 1\) 3) \(p_{3}x + q_{3}y = 1\)

Step 2: Express the concurrency condition using determinants

We can write the concurrency condition using the Cramer's rule with determinants as follows: \(\begin{vmatrix} p_{1} & q_{1} & 1 \\ p_{2} & q_{2} & 1 \\ p_{3} & q_{3} & 1 \end{vmatrix} = 0\)

Step 3: Expand the determinant

Let's expand the determinant using the first row: \( \begin{vmatrix} q_{2} & 1 \\ q_{3} & 1 \end{vmatrix} p_{1} - \begin{vmatrix} p_{2} & 1 \\ p_{3} & 1 \end{vmatrix} q_{1} + \begin{vmatrix} p_{2} & q_{2} \\ p_{3} & q_{3} \end{vmatrix} = 0\) Now simplify it: \((q_{2} - q_{3})p_{1} - (p_{2} - p_{3})q_{1} + (p_{2}q_{3} - p_{3}q_{2}) = 0\)

Step 4: Rewrite the simplified equation

Rearrange the equation: \(p_{1}(q_{2}-q_{3}) + q_{1}(p_{3}-p_{2}) = p_{2}q_{3} - p_{3}q_{2}\) Notice that this equation can be written as the determinant of the points (p1, q1), (p2, q2), and (p3, q3): \(\begin{vmatrix} p_{1} & q_{1} & 1 \\ p_{2} & q_{2} & 1 \\ p_{3} & q_{3} & 1 \end{vmatrix} = p_{1}(q_{2}-q_{3}) - q_{1}(p_{2}-p_{3}) + (p_{2}q_{3} - p_{3}q_{2})\)

Step 5: Conclude the proof

Since the determinant of the points (p1, q1), (p2, q2), and (p3, q3) is equal to the determinant expressing the concurrency condition, and as the three lines are concurrent, their determinant is zero. Thus, the determinant of the points (p1, q1), (p2, q2), and (p3, q3) is also zero, which implies that these points are collinear.

Key Concepts

Cramer's Rule

Cramer's rule is a theorem in linear algebra, which gives an explicit expression for the solution of a system of linear equations with as many equations as unknowns, provided the system has a unique solution. It states that a unique solution exists if and only if the determinant of the system's matrix is non-zero.

Applying Cramer's rule involves using determinants to solve for each variable. For a system of equations given as:

• \(a_1x + b_1y = c_1\)
• \(a_2x + b_2y = c_2\)

the solutions for \(x\) and \(y\) would be:

• \(x = \frac{\begin{vmatrix} c_1 & b_1 \ c_2 & b_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1 \ a_2 & b_2 \end{vmatrix}}\)
• \(y = \frac{\begin{vmatrix} a_1 & c_1 \ a_2 & c_2 \end{vmatrix}}{\begin{vmatrix} a_1 & b_1 \ a_2 & b_2 \end{vmatrix}}\)

In the given exercise, Cramer's rule is used to set up the condition for the concurrency of three lines. If they are concurrent, they satisfy the single point solution set by Cramer's rule, where the determinant composed of the coefficients of the lines is equal to zero.

Determinants

A determinant is a scalar value that can be computed from the elements of a square matrix. It provides important information about the matrix, such as whether it is invertible and the volume scaling factor for the linear transformation that the matrix represents. The determinant of a 2x2 matrix \(A\) is given by:

• \(det(A) = \begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc\)

For a 3x3 matrix \(B\):

• \(det(B) = \begin{vmatrix} e & f & g \ h & i & j \ k & l & m \end{vmatrix}\)

The computation can be more involved, following specific rules and expansion along rows or columns.

In our example, the determinant is used to establish the concurrency condition. The determinant must be zero for the lines to be concurrent. This is a critical concept because it links the algebraic properties of the linear equations with the geometric properties of the lines represented by these equations.

Collinear Points

Points are said to be collinear if they lie on the same straight line. In the two-dimensional plane, collinearity can be tested using various methods, with one involving determinants. The determinant composed of the coordinates of the points can indicate collinearity. If the determinant equals zero, this implies that the area of the triangle formed by the points is zero; therefore, they must be on the same line.

For example, points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) are collinear if the determinant of their coordinates is zero:

• \(\begin{vmatrix} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{vmatrix} = 0\)

The exercise improvement advice points out that understanding the geometric interpretation of this determinant as the area of a triangle and linking it to the definition of collinear points can be particularly helpful for students. This visualization aids in grasping why a zero determinant corresponds to collinearity of points.