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Chapter 2: Problem 34

Use the given conditions to write an equation for each line in point slope form and slope-intercept form. Passing through \((-2,-5)\) and \((6,-5)\)

Short Answer

Expert verified
The equation of the line in point-slope form is \(y + 5 = 0\), and in slope intercept form is \(y = -5\).

Step by step solution

01

Calculate the Slope

The slope of the line passing through two given points \((-2,-5)\) and \((6,-5)\) is given by \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Here, \((x_1, y_1) = (-2, -5)\) and \((x_2, y_2) = (6, -5)\). So, \(m = \frac{-5 - (-5)}{6 - (-2)} = \frac{0}{8} = 0\). Thus, the slope of the line is 0.
02

Point-Slope Form

Now that the slope is known, use the point-slope form equation, \(y - y_1 = m(x - x_1)\). Substituting \((x_1, y_1) = (-2, -5)\) and \(m = 0\) into this equation, you get \(y + 5 = 0 * (x + 2)\) which simplifies to \(y + 5 = 0\). Hence, the point-slope form of the equation is \(y + 5 = 0\).
03

Slope-Intercept Form

Finally, the slope-intercept form equation \(y = mx + b\) is used. Substituting \(m = 0\) into the equation, it becomes \(y = 0*x + b\). To find \(b\), pick one of the points and substitute the x and y values. Using point \((-2,-5)\), the equation is \(-5 = 0 * -2 + b\), which simplifies to \(-5 = b\). Hence, the slope-intercept form of the equation is \(y = -5\).

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Key Concepts

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

Point-Slope Form
Point-slope form is a useful way to express the equation of a straight line. It comes in handy especially when you know a point on the line and its slope. The formula is given by:
  • \( y - y_1 = m(x - x_1) \)
where \((x_1, y_1)\) is a point on the line and \(m\) is the slope. In our example, we used the point \((-2, -5)\) and a slope of 0. Plug these into the formula and you get the equation in point-slope form:
  • \( y + 5 = 0 \)
The point-slope form helps us understand that with a slope of 0, the line is horizontal, and the y-value doesn't change.
Slope-Intercept Form
Slope-intercept form is perhaps the most popular way to represent a linear equation. It is easy to grasp because it directly reveals the slope and the y-intercept of the line. The general formula is:
  • \( y = mx + b \)
Here, \(m\) represents the slope and \(b\) is the y-intercept. From our example, we found that the slope \(m = 0\). This makes the equation:
  • \( y = 0x + b \)
By substituting any of the given points, say \((-2, -5)\), we find that \(b = -5\). Therefore, the slope-intercept form is simple and tells us:
  • \( y = -5 \)
This demonstrates that the line is parallel to the x-axis at \(y = -5\).
Calculate Slope
Calculating the slope of a line is a crucial skill in understanding linear equations. The slope measures how steep a line is and is calculated as:
  • \( m = \frac{y_2 - y_1}{x_2 - x_1} \)
This formula takes two points, \((x_1, y_1)\) and \((x_2, y_2)\), and finds the difference in their y-values divided by the difference in their x-values. For example, with the points \((-2, -5)\) and \((6, -5)\), the slope is:
  • \( \frac{-5 - (-5)}{6 - (-2)} = \frac{0}{8} = 0 \)
A slope of 0 indicates a horizontal line, where the y-value does not change no matter the x-value.
Linear Equations
Linear equations create straight lines when graphed. They follow the general form:
  • \( ax + by + c = 0 \)
or can be rearranged to forms such as point-slope or slope-intercept. These equations relate two variables, typically x and y, in a manner showing consistent rate of change, or slope. A horizontal line, like in our recent example, results from a slope of 0 and appears as a constant y-value equation:
  • \( y = -5 \)
Linear equations are foundational in algebra and crucial for understanding more complex mathematical concepts later on.

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Most popular questions from this chapter

If a function is defined by an equation, explain how to find its domain.

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. If \(f(x)=\sqrt{x}\) and \(g(x)=2 x-1,\) then \((f \circ g)(5)=g(2)\)

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}+x+y-\frac{1}{2}=0$$

Give the center and radius of the circle described by the equation and graph each equation. Use the graph to identify the relation鈥檚 domain and range. $$(x+4)^{2}+(y+5)^{2}=36$$

Find a. \((f \circ g)(x)\) b. \((g \circ f)(x)\) c. \((f \circ g)(2)\) d. \((g \circ f)(2)\) $$f(x)=x^{2}+2, g(x)=x^{2}-2$$
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