Problem 79
Question
Find the value of \(y\) if the line through the two given points is to have the indicated slope. \((3, y)\) and \((1,4), m=-3\)
Step-by-Step Solution
Verified Answer
The value of \(y\) is 2.
1Step 1: Define the points and slope
According to the problem, the points are \((3, y)\) and \((1,4)\) and the slope \(m\) is -3.
2Step 2: Apply the slope formula
The formula for the slope of a line given two points is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). In this case, identify \((3, y)\) as \((x_1, y_1)\) and \((1, 4)\) as \((x_2, y_2)\). Plug the values into the formula \(-3 = \frac{4 - y}{1 - 3}\).
3Step 3: Solve for \(y\)
The equation \(-3 = \frac{4 - y}{1 - 3}\) simplifies to \(-3 = \frac{4-y}{-2}\), which then simplifies to -3 = \(-(4-y)\). Solving this equation gives \(y = 2\).
Key Concepts
Finding the Value of YSlope FormulaLinear Equations
Finding the Value of Y
Understanding how to find the value of 'y' in the context of lines on a coordinate plane is crucial. When given two points that a line passes through, it might be necessary to determine one of the coordinates if it is not already known. For example, if you have a line passing through the points (3, y) and (1,4), and you know the slope (-3), you would substitute the known values into the slope formula to find the missing 'y' value.
Let's look at this process more closely. With the slope formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\), by replacing (x_2, y_2) with (1, 4) and (x_1, y_1) with (3, y), we arrive at \(-3 = \frac{4 - y}{1 - 3}\). Simplifying the denominator (-2) and solving for 'y' by doing the appropriate algebraic manipulations, leads us to the solution \(y = 2\). In this way, missing coordinates can be found comfortably by following algebraic principles.
Let's look at this process more closely. With the slope formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\), by replacing (x_2, y_2) with (1, 4) and (x_1, y_1) with (3, y), we arrive at \(-3 = \frac{4 - y}{1 - 3}\). Simplifying the denominator (-2) and solving for 'y' by doing the appropriate algebraic manipulations, leads us to the solution \(y = 2\). In this way, missing coordinates can be found comfortably by following algebraic principles.
Slope Formula
The slope of a line is a measure of how steep the line is and is a foundational concept in algebra. In mathematical terms, slope is defined as the rise over the run, or the change in 'y' over the change in 'x'. The formula for slope when considering two points \((x_1, y_1)\) and \((x_2, y_2)\) on a line is expressed as \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
The formula is quite straightforward when you remember that you are simply subtracting the 'y' values of the points and dividing by the difference in the 'x' values of the same points. Remembering that slope is usually represented by 'm', it becomes easier to use this formula to find a line's slope—or in our case, a missing coordinate that fits a specific slope.
The formula is quite straightforward when you remember that you are simply subtracting the 'y' values of the points and dividing by the difference in the 'x' values of the same points. Remembering that slope is usually represented by 'm', it becomes easier to use this formula to find a line's slope—or in our case, a missing coordinate that fits a specific slope.
Linear Equations
Linear equations form the bedrock of algebra and describe two-dimensional lines. They take the general form \(y = mx + b\), where 'm' represents the slope, and 'b' stands for the y-intercept, the point at which the line crosses the y-axis. Understanding how to manipulate these equations is essential when dealing with graphing lines, finding slopes, or determining points on a line.
When working with linear equations, one of the goals might be solving for 'y', like in the exercise above. To solve for 'y', rearrange the equation so that 'y' is by itself on one side. This process typically involves isolating the 'y' term through inverse operations like addition or subtraction and division or multiplication. Once 'y' is isolated, the equation can be used to make predictions, analyze the slope, and draw comparisons between different lines.
When working with linear equations, one of the goals might be solving for 'y', like in the exercise above. To solve for 'y', rearrange the equation so that 'y' is by itself on one side. This process typically involves isolating the 'y' term through inverse operations like addition or subtraction and division or multiplication. Once 'y' is isolated, the equation can be used to make predictions, analyze the slope, and draw comparisons between different lines.
Other exercises in this chapter
Problem 79
Express the given function \(h\) as a composition of two functions \(f\) and \(g\) so that \(h(x)=(f \circ g)(x)\). $$h(x)=|2 x-5|$$
View solution Problem 79
Begin by graphing the square root function, \(f(x)=\sqrt{x} .\) Then use transformations of this graph to graph the given function. $$g(x)=2 \sqrt{x+2}-2$$
View solution Problem 80
Does \((x-3)^{2}+(y-5)^{2}=0\) represent the equation of a circle? If not, describe the graph of this equation.
View solution Problem 80
Use a graphing utility to graph the function. Use the graph to determine whether the function has an inverse that is a function (that is, whether the function i
View solution