Problem 24
Question
Write the point-slope form of the equation of the line satisfying each of the conditions in Exercises \(1-28 .\) Then use the point-slope form of the equation to write the slope-intercept form of the equation. Passing through \((-2,-5)\) and \((6,-5)\)
Step-by-Step Solution
Verified Answer
The equation of the line passing through points (-2,-5) and (6,-5) is \(y = -5\).
1Step 1: Calculate the Slope
The first step is to determine the slope of the line. The slope 'm' is calculated as difference in y-coordinates divided by difference in x-coordinates, i.e., \((y_2-y_1)/(x_2-x_1)\). For the points (-2,-5) and (6,-5), the calculation becomes \((-5-(-5))/(6-(-2))\). But, as the y-coordinates of both points are same, so the slope (m) is 0 because the denominator doesn't affect the slope.
2Step 2: Substitute in the Point-Slope Form
After determining the slope, it should be substituted into the point-slope form of the line equation formula, which is \(y - y_1 = m(x - x_1)\). In this formula, \(y_1\) and \(x_1\) are coordinates of any point which the line passes through. Substituting \(m=0\), \(x_1 = -2\) and \(y_1 = -5\) we get \(y - (-5) = 0 (x + 2)\). This simplifies to \(y + 5 = 0\).
3Step 3: Convert to Slope-Intercept Form
The final step is to convert the point-slope form into the slope-intercept form \(y = mx + b\). In doing so, we simplify the equation from Step 2 to bring it into this form, in this case, \(y = -5\).
Key Concepts
AlgebraSlope CalculationSlope-Intercept Form
Algebra
Algebra is a branch of mathematics dealing with symbols and the rules for manipulating these symbols; it is a unifying thread of almost all of mathematics. It allows us to express quantities and their relationships to each other in formulae and equations, which can be solved to find unknowns. In the context of plotting lines on a coordinate plane, algebra is used to derive different forms of equations that represent lines. These equations are crucial in understanding the slope, direction, and position of the line in relation to the origin.
When you work with the point-slope form of a line's equation, you are applying algebraic methods to find a linear equation that uniquely represents a line given a point and a slope. This process involves substituting values into the point-slope form and, if necessary, simplifying it into the slope-intercept form, which is a frequent requirement for various algebraic applications.
When you work with the point-slope form of a line's equation, you are applying algebraic methods to find a linear equation that uniquely represents a line given a point and a slope. This process involves substituting values into the point-slope form and, if necessary, simplifying it into the slope-intercept form, which is a frequent requirement for various algebraic applications.
Slope Calculation
Slope calculation is a fundamental concept in algebra that determines the steepness, incline, or decline of a line in a coordinate plane. The slope is identified by the symbol 'm' and can be found using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\), where \(x_1, y_1\) and \(x_2, y_2\) are the coordinates of two distinct points on the line.
In the exercise, you will notice that both points lie on the same horizontal level, which means the change in the y-coordinates \( \Delta y\) is zero. As a result, the slope of such a line is 0, indicating that the line is horizontal. Understanding this fundamental concept helps in not just graphing lines but also in solving systems of linear equations, finding parallel and perpendicular lines, and analyzing real-world linear relationships.
In the exercise, you will notice that both points lie on the same horizontal level, which means the change in the y-coordinates \( \Delta y\) is zero. As a result, the slope of such a line is 0, indicating that the line is horizontal. Understanding this fundamental concept helps in not just graphing lines but also in solving systems of linear equations, finding parallel and perpendicular lines, and analyzing real-world linear relationships.
Slope-Intercept Form
The slope-intercept form \(y = mx + b\) is widely used in algebra to describe the equation of a straight line, where 'm' stands for the slope and 'b' denotes the y-intercept. The y-intercept is the point where the line crosses the y-axis. By rearranging the point-slope equation into the slope-intercept form, you can readily see what the slope (\(m\)) of the line is and where the line crosses the y-axis (\(b\)).
In the given exercise, after determining the slope is zero, we plug the values into the point-slope form and rewrite it to the slope-intercept form, revealing the y-intercept directly. In our case, there is no x-term since the slope is zero, indicating a horizontal line crossing the y-axis at \(-5\). The beauty of the slope-intercept form is its simplicity in graphing as well as its practical use in comparing different linear equations quickly.
In the given exercise, after determining the slope is zero, we plug the values into the point-slope form and rewrite it to the slope-intercept form, revealing the y-intercept directly. In our case, there is no x-term since the slope is zero, indicating a horizontal line crossing the y-axis at \(-5\). The beauty of the slope-intercept form is its simplicity in graphing as well as its practical use in comparing different linear equations quickly.
Other exercises in this chapter
Problem 23
Use intercepts and a checkpoint to graph each equation. $$6 x-9 y=18$$
View solution Problem 23
Plot the given point in a rectangular coordinate system. $$\left(0,-\frac{5}{2}\right)$$
View solution Problem 24
In Exercises \(13-26,\) begin by solving the linear equation for \(y .\) This will put the equation in slope-intercept form. Then find the slope and the \(y\) -
View solution Problem 24
Determine whether the distinct lines through each pair of points are parallel. $$(2,4)\( and \)(6,1) ;(-3,1)\( and \)(1,-2)$$
View solution