Problem 44 - College Algebra with Modeling and Visualization

Question

If a line passes through the points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right),\) then an equation of this line can be found by calculating the determinant. $$ \operatorname{det}\left[\begin{array}{lll} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{array}\right]=0 $$ Find the standard form ax \(+b y=c\) of the line passing through the given points. \((5,1)\) and \((2,-2)\)

Step-by-Step Solution

Answer

The equation is \(x - y = 4\).

Step 1: Set Up the Determinant

To find the equation of the line, we need to set up the determinant with the given points \((5,1)\) and \((2,-2)\). Insert the values into the determinant matrix:\[\operatorname{det}\begin{bmatrix}x & y & 1 \5 & 1 & 1 \2 & -2 & 1\end{bmatrix}=0\]

Step 2: Calculate the Determinant

Calculate the determinant of the matrix. Expand it as follows:\[\operatorname{det} = x(1 - (-2)) - y(5 - 2) + 1((-10) - 2) = 0\]This simplifies to:\[3x - 3y - 12 = 0\]

Step 3: Simplify the Equation

Rearrange the equation from the determinant to obtain the standard form \(ax + by = c\):\[3x - 3y = 12\]Now, divide the whole equation by 3 to simplify it:\[x - y = 4\]This is the standard form of the equation of the line.

Key Concepts

Determinant

The determinant is a special value that can be computed from a matrix. In our exercise, we use a 3x3 matrix to express points that a line passes through. The determinant helps us determine when three points form a collinear line. This is because a line through these two points can be described using the matrix's determinant set to zero.
To calculate the determinant for the given line, we use the expanded formula:

• For a 3x3 matrix \[\begin{vmatrix} x & y & 1 \ x_{1} & y_{1} & 1 \ x_{2} & y_{2} & 1 \end{vmatrix}\] we calculate as follows:
• Multiply \((x\cdot (y_1-y_2))\), subtract \((y \cdot (x_1-x_2))\), and add \(1\cdot (x_1 \cdot y_2 - x_2 \cdot y_1)\).

This determinant will always equal zero for a line passing through those points, which lets us write the line's equation.

Standard Form

The standard form of a line's equation is expressed as \(ax + by = c\). This form is beneficial because its coefficients \(a\), \(b\), and \(c\) give intuitive insights into the properties of the line, such as slope and intercepts.
To convert into the standard form, begin by simplifying the determinant-derived expression. For instance, from \[3x - 3y - 12 = 0\],transform it into the expression:

• First, reorganize: \(3x - 3y = 12\). This puts x and y terms on one side and the constant on the other.
• Then, divide through by the common factor of 3 to simplify: \(x - y = 4\).

This gives the standard form, indicating simplification and rearrangement are key steps.

Coordinate Geometry

Coordinate Geometry, also known as analytic geometry, combines algebra and geometry using a coordinate system. It allows for precise graphs and calculations of geometric concepts, like lines, using coordinates.
Relating this to our line equation:

• Points \((x_1, y_1)\) and \((x_2, y_2)\) describe the line's path.
• Analytic methods let us derive equations (like the line's) directly from these coordinates.
• This approach provides deeper insights: distance formulas, midpoint, and slope calculations become straightforward, as coordinates define geometric objects within a plane.

The mixed use of algebra and geometry in coordinate systems simplifies finding line equations and understanding geometric relationships. This is fundamental in fields ranging from engineering to computer graphics.