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Chapter 1: Problem 46

\(45-50\) Find an expression for the function whose graph is the given curve. $$\begin{array}{l}{\text {The line segment joining the points } (-5,10) and (7,-10)}\end{array}$$

Short Answer

Expert verified
The equation of the line segment is \(y = -\frac{5}{3}x + \frac{5}{3}\).

Step by step solution

01

Understand the Problem

You need to find the equation of the line segment that joins two points: \((-5, 10)\) and \((7, -10)\). This involves finding the linear equation of the form \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept.
02

Calculate the Slope

The slope \(m\) of a line through two points \((x_1, y_1)\) and \((x_2, y_2)\) is calculated as \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Using the points \((-5, 10)\) and \((7, -10)\), the slope is \(m = \frac{-10 - 10}{7 - (-5)} = \frac{-20}{12} = -\frac{5}{3}\).
03

Use Point-Slope Form

Use the point-slope form of the equation of a line: \(y - y_1 = m(x - x_1)\). Choose one of the points, say \((-5, 10)\). Substitute \(m = -\frac{5}{3}\), \(x_1 = -5\), and \(y_1 = 10\) into the equation: \(y - 10 = -\frac{5}{3}(x + 5)\).
04

Simplify to Slope-Intercept Form

Solve for \(y\) to get the line in the form \(y = mx + c\):\[\begin{align*}y - 10 &= -\frac{5}{3}(x + 5) \y - 10 &= -\frac{5}{3}x - \frac{25}{3} \y &= -\frac{5}{3}x - \frac{25}{3} + 10 \y &= -\frac{5}{3}x - \frac{25}{3} + \frac{30}{3} \y &= -\frac{5}{3}x + \frac{5}{3}\end{align*}\]
05

Verify the Line Passes Through Both Points

Substitute \(x = -5\) and \(x = 7\) into the equation \(y = -\frac{5}{3}x + \frac{5}{3}\) to check: For \((-5, 10)\):\[\begin{align*}y &= -\frac{5}{3}(-5) + \frac{5}{3} \y &= \frac{25}{3} + \frac{5}{3} \y &= \frac{30}{3} = 10\end{align*}\]For \((7, -10)\):\[\begin{align*}y &= -\frac{5}{3}(7) + \frac{5}{3} \y &= -\frac{35}{3} + \frac{5}{3} \y &= -\frac{30}{3} = -10\end{align*}\]Both points satisfy the equation, confirming that the line is correct.

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Key Concepts

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

Slope
The slope is a measure of how steep a line is. It tells us how much the line rises (or falls) as we move from left to right on a graph. Imagine you're walking up or down a hill; the slope is a bit like the steepness of that hill. For a line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \), the slope \( m \) is calculated using the formula: \[ m = \frac{y_2-y_1}{x_2-x_1} \].
This gives us a number that can be:
  • Positive, which means the line goes upwards.
  • Negative, which means the line goes downwards.
  • Zero, which means the line is flat.
  • Undefined, which happens for vertical lines.
The larger the absolute value of the slope, the steeper the line. In our exercise, when calculated using the points \( (-5, 10) \) and \( (7, -10) \), the slope is found to be \[ -\frac{5}{3} \], indicating a downward slope.
Y-Intercept
The y-intercept is the point where the line crosses the y-axis of a graph. In simpler terms, it's what the value of \( y \) is when \( x \) is zero. This is an important number since it gives us an easy starting point for graphing a line. For a line in the slope-intercept form, \( y = mx + c \), the \( c \) represents the y-intercept.
In the exercise's final solution, the line's equation is \( y = -\frac{5}{3}x + \frac{5}{3} \). Here, the y-intercept is \( \frac{5}{3} \). This means when \( x \) equals zero, \( y \) is \( \frac{5}{3} \). You can imagine it as a starting point for the line on a graph.
Point-Slope Form
The point-slope form of a line equation is useful when we know one point on the line and the slope. The form is written as \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \( (x_1, y_1) \) is the given point. Let's break down why this form is particularly handy:
  • You only need one point on the line and the slope.
  • It can be easily rearranged to other forms, like slope-intercept.
  • It's straightforward to plug numbers in and solve.
To use the point-slope form in our exercise, we chose the point \( (-5, 10) \) and used the calculated slope \( -\frac{5}{3} \). The equation derived was \( y - 10 = -\frac{5}{3} (x + 5) \). From here, it's simple to rearrange to find the slope-intercept form, which we did by solving for \( y \). This process shows how flexible and adaptable the point-slope form can be.

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