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Show that the equation \(\left|\begin{array}{lll}x & y & 1 \\ x_{1} & y_{1} & 1 \\ 1 & m & 0\end{array}\right|=0\) represents a line that has slope \(m\) and passes through the point \(\left(x_{1}, y_{1}\right)\)

Short Answer

Expert verified
The equation represents a line with slope \( m \) passing through \((x_1, y_1)\).

Step by step solution

01

Expand the Determinant

To show that the given determinant represents a line with slope \( m \), start by expanding the determinant \( \left|\begin{array}{lll}x & y & 1 \ x_{1} & y_{1} & 1 \ 1 & m & 0\end{array}\right| \). Expanding this determinant, the expression will be a combination of the minors: \( x(y_{1} \cdot 0 - 1 \cdot m) - y(x_{1} \cdot 0 - 1 \cdot 1) + 1(x_{1}m - y_{1}\cdot 1) \).
02

Simplify the Expression

Simplify the expanded expression: \(-xm - y(-1) + x_1 m - y_1 = -xm + y + x_1 m - y_1 \). Simplifying further, combine similar terms to arrive at: \( y = xm - x_1 m + y_1 \).
03

Arrange the Equation

Rearrange the simplified equation \( y = xm - x_1 m + y_1 \) to the standard line equation form: \( y - y_1 = m(x - x_1) \). This equation represents a line in point-slope form, indicating a line with slope \( m \) passing through the point \((x_1, y_1)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Slope of a Line
The slope of a line is a measure of how steep the line is, and it indicates the relationship between the change in the vertical direction (rise) and the change in the horizontal direction (run) between two points on the line. Mathematically, the slope is defined as:\[m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}\]where
  • \((x_1, y_1)\) and \((x_2, y_2)\) are coordinates of two distinct points on the line,
  • \(\Delta y\) is the change in the y-coordinates, and
  • \(\Delta x\) is the change in the x-coordinates.
The slope tells us many things about a line:
  • If \(m > 0\), the line is sloping upwards from left to right.
  • If \(m < 0\), the line is sloping downwards.
  • If \(m = 0\), the line is horizontal, and
  • If the line is vertical, the slope is undefined because \(\Delta x = 0\).
In the context of the given equation, the slope \(m\) is shown through the process of expanding the determinant. This indicates the line's inclination compared to a horizontal line.
Point-Slope Form of a Line Equation
The point-slope form of a line is a handy way to describe a straight line when you know one point on the line and its slope. The equation is given by:\[y - y_1 = m(x - x_1)\]where
  • \((x_1, y_1)\) is a known point on the line, and
  • \(m\) is the slope of the line.
This form is especially useful because it directly incorporates the slope and a specific point, making it easier to visualize and understand the line's behavior.
In our exercise, after expanding and simplifying the determinant, we end up with this same form, illustrating that the expression represents a line through the given point with the specified slope. This shows how determinants can be used to derive the equation of a line efficiently, by simplifying geometric properties into algebraic expressions.
Expanding Determinants to Formulate Equations
Determinants offer a powerful method to handle equations in matrix form. By expanding determinants, as in this exercise, we reformulate a matrix into a more comprehensible algebraic expression.
The process of expanding the determinant \(\left|\begin{array}{lll}x & y & 1 \ x_{1} & y_{1} & 1 \ 1 & m & 0\end{array}\right|\) involves computing the product of its diagonal terms and then subtracting the product of other minor matrices. This results in:\[x(y_1 \cdot 0) - xm - y(x_1 \cdot 0 + 1) + 1(x_1 \cdot m - y_1 \cdot 1)\]through careful expansion of each row.
Upon simplification, this expression becomes \(y - y_1 = m(x - x_1)\), which, as we discovered, describes the point-slope form of a line.
  • Expanding determinants can simplify complex matrix expressions into a form that is easier to analyze.
  • This technique helps relate the properties of the matrix (or equation) to geometric concepts such as lines and their slopes.
Thus, by understanding determinants and their expansions, you gain a useful tool for translating geometric scenarios into algebraic language.

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Most popular questions from this chapter

Show that $$\left|\begin{array}{lll} 1 & a & a^{2} \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right|=\left(a^{3}-1\right)^{2}$$

Use a graphing utility to compute the matrix products. $$\left(\begin{array}{rr} 1.03 & 2.1 \\ -0.45 & 3.09 \end{array}\right)\left(\begin{array}{rr} 2.33 & 4.17 \\ 0 & -1.24 \end{array}\right)$$

So the input (3,5) yields an output of \(\sqrt{2} .\) We define the do- main for this function just as we did in Chapter 3: The domain is the set of all inputs that yield real-number outputs. For instance, the ordered pair (1,4) is not in the domain of the function we have been discussing, because (as you should check for yourself\() f(1,4)=\sqrt{-1},\) which is not a real number. We can determine the domain of the function in equation ( 1 ) by requiring that the quantity under the radical sign be non negative. Thus we require that \(2 x-y+1 \geq 0\) and, consequently, \(y \leq 2 x+1\) (Check this.) The following figure shows the graph of this inequality; the domain of our function is the set of ordered pairs making up the graph. In Exercises follow a similar procedure and sketch the domain of the given function. (Graph cant copy) $$f(x, y)=\sqrt{x+y+2}$$

A parabola \(y=a x^{2}+b x+c\) passes through the three points \((1,-2),(-1,0),\) and (2,3) (a) Write down a system of three linear equations that must be satisfied by \(a, b,\) and \(c\) (b) Solve the system in part (a). (c) With the values for \(a, b,\) and \(c\) that you found in part (b), use a graphing utility to draw the parabola \(y=a x^{2}+b x+c .\) Find a viewing rectangle that seems to confirm that the parabola indeed passes through the three given points. (d) Another way to check your result in part (b): Apply the quadratic regression option on a graphing utility after entering the three given data points \((1,-2),(-1,0),\) and \((2,3) .\) (This is a valid check because, in general, three noncollinear points determine a unique parabola.)

By expanding the determinant \(\left|\begin{array}{ccc}k a & k b & k c \\ d & e & f \\ g & h & i\end{array}\right|\) along its first row, show that it is equal to \(k\left|\begin{array}{lll}a & b & c \\ d & e & f \\ g & h & i\end{array}\right|\)

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