Problem 28 - The Calculus with Analytic Geometry

Question

Let \(l_{1}\) be the line having the equation \(A_{1} x+B_{1} y+C_{1}=0\), and let \(l_{2}\) be the line having the equation \(A_{2} x+B_{2} y+C_{2}=0\). If \(l_{1}\) is not parallel to \(l_{2}\) and if \(k\) is any constant, the equation $$ A_{1} x+B_{1} y+C_{1}+k\left(A_{2} x+B_{2} y+C_{2}\right)=0 $$ represents an unlimited number of lines. Prove that each of these lines contains the point of intersection of \(l_{1}\) and \(l_{2}\).

Step-by-Step Solution

Answer

Each line in the family of lines given by the equation contains the intersection point of \(l_1\) and \(l_2\).

Step 1 - Find the intersection point of the lines

To find the point of intersection of the lines, solve the system of equations given by the two lines: ewline\(A_{1} x + B_{1} y + C_{1} = 0\) and \(A_{2} x + B_{2} y + C_{2} = 0\). ewlineSet these equations equal to each other and solve for \(x\) and \(y\).

Step 2 - Solve system of equations

Solving the system involves using either the substitution method or the elimination method. For instance, using elimination: ewlineMultiplying the first equation by \(B_{2}\) and the second equation by \(B_{1}\) gives: ewline\[A_{1}B_{2} x + B_{1}B_{2} y + C_{1}B_{2} = 0\] ewline and ewline \[A_{2}B_{1} x + B_{2}B_{1} y + C_{2}B_{1} = 0\]. ewlineThen, subtract one equation from the other to eliminate \(y\), solving for \(x\).

Step 3 - Find the point (x_0, y_0)

Once \(x\) is found, substitute it back into one of the original equations to solve for \(y\). This gives the point of intersection \((x_0, y_0)\) of \(l_{1}\) and \(l_{2}\).

Step 4 - Construct the generalized line equation

Consider the equation given in the problem: ewline\(A_{1} x + B_{1} y + C_{1} + k (A_{2} x + B_{2} y + C_{2}) = 0\). ewlineWe need to show this passes through \((x_0, y_0)\).

Step 5 - Substitute the intersection point

Substitute \((x_0, y_0)\) into the generalized equation: ewline\[A_{1} x_0 + B_{1} y_0 + C_{1} + k (A_{2} x_0 + B_{2} y_0 + C_{2}) = 0\]. ewlineSince \((x_0, y_0)\) satisfies both original line equations, it must satisfy: ewline\[A_{1} x_0 + B_{1} y_0 + C_{1} = 0\] and ewline \[A_{2} x_0 + B_{2} y_0 + C_{2} = 0\]. ewlineThus, the generalized line equation will satisfy ewline \[0 + k \times 0 = 0\].

Step 6 - Conclusion

Since \((x_0, y_0)\) satisfies the equation for any value of \(k\), the conclusion is that all such lines contain the point of intersection of \(l_1\) and \(l_2\). Therefore, the statement is proven.

Key Concepts

system of linear equations

A system of linear equations consists of two or more linear equations involving the same set of variables. In the context of the given problem, the system of equations is formed by the two lines:
\(A_{1} x + B_{1} y + C_{1} = 0\) and \(A_{2} x + B_{2} y + C_{2} = 0\).
The solution to this system is the point \((x_0, y_0)\) where the two lines intersect. This point represents the coordinates that satisfy both equations simultaneously.

To find the intersection point, you typically solve the system using methods like elimination or substitution. Understanding how to solve systems of linear equations is crucial for locating the intersection of lines in analytic geometry.

elimination method

The elimination method is a technique for solving systems of linear equations. It involves manipulating the equations to eliminate one variable, making it easier to solve for the other variable.

For example, to solve the equations:
\(A_{1} x + B_{1} y + C_{1} = 0\) and \(A_{2} x + B_{2} y + C_{2} = 0\), you can multiply each equation by a suitable value to align coefficients and then subtract one equation from the other to eliminate a variable.

In the provided solution, the first equation is multiplied by \(B_{2}\) and the second by \(B_{1}\), resulting in:
\(A_{1} B_{2} x + B_{1} B_{2} y + C_{1} B_{2} = 0\)
and
\(A_{2} B_{1} x + B_{2} B_{1} y + C_{2} B_{1} = 0\)

By subtracting these two equations, the \(y\) terms are eliminated, making it possible to solve for \(x\). Once \(x\) is found, substituting it back into one of the original equations will give the value of \(y\).

substitution method

The substitution method is another technique for solving systems of linear equations. This method involves solving one of the equations for one variable and then substituting that expression into the other equation.

For the system:
\(A_{1} x + B_{1} y + C_{1} = 0\) and \(A_{2} x + B_{2} y + C_{2} = 0\), you can solve the first equation for \(y\):
\(y = -\frac{A_{1} x + C_{1}}{B_{1}}\).

Then, substitute this expression for \(y\) into the second equation:
\(A_{2} x + B_{2} \left( -\frac{A_{1} x + C_{1}}{B_{1}} \right) + C_{2} = 0\).

This substitution results in a single equation with one variable, which can be solved for \(x\). After finding \(x\), the value can be substituted back into the first equation to determine \(y\). This point \((x_0, y_0)\) represents the intersection of the two lines in the system.

analytic geometry

Analytic geometry, also known as coordinate geometry, is the study of geometry using a coordinate system. This branch of mathematics uses algebraic equations to represent geometric shapes and solve geometric problems.

In the context of the provided exercise, analytic geometry helps us understand the relationship between different lines through algebraic equations. The lines \(l_1\) and \(l_2\) are defined by their respective equations:
\(A_{1} x + B_{1} y + C_{1} = 0\) and \(A_{2} x + B_{2} y + C_{2} = 0\).

These equations describe straight lines in a coordinate plane, and analytic geometry provides the tools to find their intersection. By solving these equations together, we find the point \((x_0, y_0)\) where the two lines intersect.

Furthermore, the exercise shows that the line described by the generalized equation:
\(A_{1} x + B_{1} y + C_{1} + k (A_{2} x + B_{2} y + C_{2}) = 0\)
passes through this intersection point for any value of \(k\). This demonstrates a deeper geometric relationship between the original lines through the lens of algebra, showcasing the power and beauty of analytic geometry.