Problem 65 - Contemporary Precalculus

Question

Find an equation for the line satisfying the given conditions. Find a real number \(k\) such that (3,-2) is on the line \(k x-2 y+7=0\).

Step-by-Step Solution

Answer

Answer: \(-\dfrac{11}{3}\)

Step 1: Plug the coordinates of the point into the equation

We have the equation of the line as \(kx - 2y + 7 = 0\), and the point (3, -2). Replace x with 3 and y with -2 in the equation: \(k(3) - 2(-2) + 7 = 0\)

Step 2: Solve for k

Now we have the equation: \(3k + 4 + 7 = 0\) Combine the constants and simplify the equation: \(3k = -11\) Divide by 3: \(k = -\dfrac{11}{3}\) Therefore, the real number k is \(-\dfrac{11}{3}\).

Key Concepts

Point-Slope Form

The point-slope form of a linear equation is invaluable when tackling problems related to lines and their positions on a graph. This form is especially useful when a specific point on a line and the line's slope are known. The general equation for the point-slope form is given by: \[ y - y_1 = m(x - x_1) \]Here:

• \(y_1\) and \(x_1\) are the coordinates of a given point on the line.
• \(m\) is the slope of the line.

To relate this to our problem, if we had the slope \(m\), point-slope form could directly guide us to the line's equation passing through the point (3, -2). This method, although not directly needed in our current problem, helps build understanding for when future problems introduce slopes explicitly. Moreover, knowing the point-slope form prepares you to convert line equations into different formats quickly.

Solving for Variables

In algebraic expressions and equations, finding the value of variables is a fundamental skill often called 'solving for variables.' This involves isolating the variable on one side of the equation to find its value. In our given problem, the goal was to find the real number \(k\) so that the point (3, -2) lies on the line given by the equation \(kx - 2y + 7 = 0\).Steps we undertake:

• Substitute the given point coordinates into the equation. This helps you check if the equation holds true for those coordinates.
• The substitution transforms the equation into one with a single variable.
• From there, solve algebraically for \(k\) by isolating \(k\) on one side of the equation to determine its value.

This problem-solving approach is not exclusive to linear equations. It applies broadly across mathematics whenever you need to find unknown values from equations. Mastery of this helps in algebraic problem-solving like balancing chemical equations or calculating angles in trigonometry.

Algebraic Manipulation

Algebraic manipulation refers to rearranging and simplifying equations. It helps in making equations easier to solve by performing operations, such as addition, subtraction, multiplication, division, or factoring. In our problem, algebraic manipulation is critical to finding the value of \(k\).Here's a closer look:1. **Substitute and Simplify:** After inserting the point's coordinates, the equation becomes \(3k + 4 + 7 = 0\). The next step involves simplifying these terms.

• Combine like terms, which involves adding the constants \(4\) and \(7\) together to simplify the equation further.

2. **Isolate the Variable:** By subtracting the combined constants from both sides, you work to isolate \(3k\).

• Following the operation, you obtain \(3k = -11\).

3. **Final Step:** Solve for \(k\) by dividing both sides by \(3\) to get \(k = -\dfrac{11}{3}\).The ability to manipulate and simplify equations systematically is at the heart of solving problems in algebra. Through practice, this becomes a natural way to approach equations, enabling you to work through them efficiently.