Problem 41 - 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. $$ (2,1) \text { and }(-1,4) $$

Step-by-Step Solution

Answer

The equation is \(x + y = 3\).

Step 1: Set up the Determinant

We start by writing down the determinant using the given points (2,1) and (-1,4). Substitute these points into the determinant: \[\operatorname{det}\begin{bmatrix}x & y & 1 \2 & 1 & 1 \-1 & 4 & 1\end{bmatrix} = 0\]

Step 2: Calculate the Determinant

Expand the determinant:\[\operatorname{det}\begin{bmatrix}x & y & 1 \2 & 1 & 1 \-1 & 4 & 1\end{bmatrix} = x(1 - 4) - y(2 - (-1)) + 1(2 \cdot 4 - (-1) \cdot 1)\]Simplifying, this results in:\[x(-3) - y(3) + 1(8 + 1) = -3x - 3y + 9\]

Step 3: Set the Determinant to Zero

Set the expression to equal zero, in accordance with the determinant equation.\[-3x - 3y + 9 = 0\]

Step 4: Simplify to Standard Form

Divide each term by -3 to simplify the equation and write it in standard form:\[x + y = 3\]

Key Concepts

Determinant

A determinant is a scalar value that is calculated from the elements of a square matrix. In this case, the determinant is used to find whether three points are collinear and to derive the equation of a line passing through two given points. The determinant is set up as a matrix that includes the coordinates of the points and a row of ones. Each point provides a row in the matrix:

• The first row contains variables \(x\), \(y\), and 1.
• The second and third rows contain \((x_1, y_1)\) and \((x_2, y_2)\), the coordinates of the two points.

By setting this determinant equal to zero, we ensure that the line passes exactly through the given points. To calculate a 3x3 determinant, apply the rule of Sarrus or the method of cofactoring. Simplify the expression to find the equation of the line.

Standard Form

The standard form of a linear equation is given by \(ax + by = c\), where \(a\), \(b\), and \(c\) are integers, and \(a\) should be a non-negative integer. Standard form is helpful because it provides a uniform way to express linear equations, making it easier to graph lines, identify properties like slope and intercepts, and discern relationships between multiple lines.To convert an equation into standard form:

• Make sure all variables and constants are on one side of the equation.
• Ensure that \(a\), \(b\), and \(c\) are integers. If not, multiply through by a common multiplier to eliminate fractions.
• Simplify the equation as needed, ensuring \(a\) is positive, typically by multiplying the entire equation by \(-1\) if necessary.

Hence, from our derived equation \(-3x - 3y + 9 = 0\), simplifying by multiplying through by \(-\frac{1}{3}\) gives us \(x + y = 3\), which is the line in standard form.

Linear Equations

Linear equations form the backbone of many mathematical concepts. These equations represent straight lines when graphed on a coordinate plane. The general form of a linear equation in two variables is \(ax + by = c\). Essential features include:

• The slope, which determines the angle and steepness of the line.
• The y-intercept, which is the point where the line crosses the y-axis.

A linear equation can be represented in various forms such as slope-intercept form (\(y = mx + b\)) or point-slope form (\(y - y_1 = m(x - x_1)\)). When solving problems involving linear equations, selecting the most convenient form can simplify calculations. For our problem, using a determinant allows for seamless transition into standard form, showcasing the versatility of linear equations in different mathematical processes.