1.3 Determining the Equation of a Line
- Find an equation of a line when given a point and the slope.
- Find an equation of a line when given two points.
In this section, you will learn how to work backwards from geometric information to find the algebraic equation of a line. Previously, we started with an equation and extracted information like slope and intercepts. Now, we will master the reverse process: determining the complete equation when given specific clues about the line.
An equation of a line can be written in three useful forms: the slope-intercept form, the point-slope form, and the standard form. The information you're given determines which form is most convenient to start with. Once you have any one form, you can always convert it to the others using basic algebra.
The Slope-Intercept Form
The slope-intercept form of a line is:
\[ y = mx + b \]where \(m\) represents the slope of the line, and \(b\) represents the y-intercept (the y-coordinate where the line crosses the y-axis).
This is the most commonly used form because it immediately reveals two critical features of the line: its steepness (\(m\)) and its starting height on the y-axis (\(b\)).
Problem: Find an equation of a line whose slope is \(5\), and y-intercept is \(3\).
Example 1.3.1 Solution
Since we are given the slope \(m = 5\) and the y-intercept \(b = 3\) directly, we simply substitute these values into the slope-intercept formula:
\[ y = 5x + 3 \]Problem: Find the equation of the line that passes through the point \((2, 7)\) and has slope \(3\).
Example 1.3.2 Solution
We know the slope \(m = 3\), so we can write the partial equation:
\[ y = 3x + b \]To find \(b\), we substitute the coordinates of the given point \((2, 7)\) into this equation, where \(x = 2\) and \(y = 7\):
\[ \begin{align} 7 &= 3(2) + b \\ 7 &= 6 + b \\ b &= 1 \end{align} \]Therefore, the complete equation is:
\[ y = 3x + 1 \]Problem: Find an equation of the line that passes through the points \((-1, 2)\) and \((1, 8)\).
Example 1.3.3 Solution
First, we calculate the slope using the two points:
\[ m = \frac{8 - 2}{1 - (-1)} = \frac{6}{2} = 3 \]Now we know the partial equation is \(y = 3x + b\). We can use either point to solve for \(b\). Using \((-1, 2)\):
\[ \begin{align} 2 &= 3(-1) + b \\ 2 &= -3 + b \\ b &= 5 \end{align} \]Thus, the equation is:
\[ y = 3x + 5 \](Check: If we substitute \((1, 8)\) into \(y = 3x + 5\), we get \(8 = 3(1) + 5\), which confirms our answer is correct.)
Problem: Find an equation of the line that has x-intercept \(3\) and y-intercept \(4\).
Example 1.3.4 Solution
An x-intercept of \(3\) means the line passes through \((3, 0)\). A y-intercept of \(4\) means the line passes through \((0, 4)\).
First, find the slope using these two points:
\[ m = \frac{4 - 0}{0 - 3} = -\frac{4}{3} \]We are explicitly told the y-intercept is \(4\), so \(b = 4\). Substituting into slope-intercept form:
\[ y = -\frac{4}{3}x + 4 \]The Point-Slope Form
The point-slope form of a line is:
\[ y - y_1 = m(x - x_1) \]where \(m\) is the slope and \((x_1, y_1)\) is any specific point that lies on the line.
To derive this formula, recall that slope \(m = \frac{y - y_1}{x - x_1}\). Multiplying both sides by \((x - x_1)\) gives us the point-slope form.
Problem: Find the point-slope form of the equation of a line that has slope \(1.5\) and passes through the point \((12, 4)\).
Example 1.3.5 Solution
Substitute \((x_1, y_1) = (12, 4)\) and \(m = 1.5\) directly into the formula:
\[ \begin{align} y - y_1 &= m(x - x_1) \\ y - 4 &= 1.5(x - 12) \end{align} \]Note: While we could simplify this to slope-intercept form by distributing and solving for \(y\), the problem specifically asks for point-slope form. In mathematics and science, leaving the answer in the requested form demonstrates that you can identify and apply the appropriate model for the given information.
The Standard Form
The standard form of a line is:
\[ Ax + By = C \]where \(A\), \(B\), and \(C\) are integers (usually with \(A \geq 0\)), and \(A\) and \(B\) are not both zero.
You can convert any slope-intercept or point-slope equation into standard form by moving all variable terms to the left side and the constant to the right side, then clearing any fractions by multiplying through by the denominator.
Problem: Using the point-slope formula, find the standard form of an equation of the line that passes through the point \((2, 3)\) and has slope \(-\frac{3}{5}\).
Example 1.3.6 Solution
Begin with point-slope form:
\[ y - 3 = -\frac{3}{5}(x - 2) \]Multiply both sides by \(5\) to eliminate the fraction:
\[ 5(y - 3) = -3(x - 2) \]Distribute:
\[ 5y - 15 = -3x + 6 \]Move all terms to the left side to set the equation equal to a constant on the right:
\[ 3x + 5y = 21 \]This is the standard form.
Problem: Find the standard form of the line that passes through the points \((1, -2)\) and \((4, 0)\).
Example 1.3.7 Solution
First, calculate the slope:
\[ m = \frac{0 - (-2)}{4 - 1} = \frac{2}{3} \]Using point-slope form with \((1, -2)\):
\[ y - (-2) = \frac{2}{3}(x - 1) \]Multiply both sides by \(3\):
\[ 3(y + 2) = 2(x - 1) \]Distribute:
\[ 3y + 6 = 2x - 2 \]Rearrange into standard form (move \(x\) and \(y\) terms to the left, constants to the right):
\[ -2x + 3y = -8 \]Or, multiplying by \(-1\) to make the \(x\)-coefficient positive:
\[ 2x - 3y = 8 \]Problem: Write the equation \(y = -\frac{2}{3}x + 3\) in standard form.
Example 1.3.8 Solution
Multiply every term by \(3\) to clear the fraction:
\[ 3y = -2x + 9 \]Add \(2x\) to both sides:
\[ 2x + 3y = 9 \]Problem: Write the equation \(3x - 4y = 10\) in slope-intercept form.
Example 1.3.9 Solution
Solve for \(y\):
\[ \begin{align} -4y &= -3x + 10 \\ y &= \frac{3}{4}x - \frac{5}{2} \end{align} \]This is now in slope-intercept form with slope \(\frac{3}{4}\) and y-intercept \(-\frac{5}{2}\).
Finding Slope by Inspection in Standard Form
There is a quick method to find the slope of a line in standard form without converting it first. If you solve \(Ax + By = C\) for \(y\), you get:
\[ y = -\frac{A}{B}x + \frac{C}{B} \]This reveals that the slope is \(-\frac{A}{B}\).
Problem: Find the slope of the following lines by inspection.
a) \(3x - 5y = 10\)
b) \(2x + 7y = 20\)
c) \(4x - 3y = 8\)
Example 1.3.10 Solution
a) Here \(A = 3\) and \(B = -5\), so:
\[ m = -\frac{3}{-5} = \frac{3}{5} \]b) Here \(A = 2\) and \(B = 7\), so:
\[ m = -\frac{2}{7} \]c) Here \(A = 4\) and \(B = -3\), so:
\[ m = -\frac{4}{-3} = \frac{4}{3} \]Problem: Find an equation of the line that passes through \((2, 3)\) and has slope \(-\frac{4}{5}\).
Example 1.3.11 Solution
Since the slope is \(-\frac{4}{5}\), we know from the inspection method that the left side of the standard form equation must be \(4x + 5y\) (because \(-\frac{A}{B} = -\frac{4}{5}\) implies \(\frac{A}{B} = \frac{4}{5}\), so we can choose \(A = 4\) and \(B = 5\)).
So we write the partial equation:
\[ 4x + 5y = c \]To find \(c\), substitute the point \((2, 3)\):
\[ \begin{align} 4(2) + 5(3) &= c \\ 8 + 15 &= c \\ c &= 23 \end{align} \]The desired equation is:
\[ 4x + 5y = 23 \]With practice, you can perform this method very quickly, jumping directly from the slope and point to the standard form equation.
Special Cases: Horizontal and Vertical Lines
A horizontal line has a slope of \(0\) and takes the form:
\[ y = b \]where \(b\) is the y-coordinate of every point on the line (the y-intercept).
A vertical line has an undefined slope and takes the form:
\[ x = a \]where \(a\) is the x-coordinate of every point on the line (the x-intercept).
Summary of Line Forms
We now have a complete toolkit for working with linear equations. Here is the summary:
| Form | Equation | Key Information Revealed |
|---|---|---|
| Slope-Intercept | \(y = mx + b\) | Slope \(m\), y-intercept \(b\) |
| Point-Slope | \(y - y_1 = m(x - x_1)\) | Slope \(m\), point \((x_1, y_1)\) |
| Standard | \(Ax + By = C\) | Easy to find both intercepts |
| Horizontal | \(y = b\) | Constant output, zero slope |
| Vertical | \(x = a\) | Fixed input, undefined slope |
You should always be able to convert from one form to another using algebraic manipulation. The form you choose depends on what information you start with and what information you need to extract.
Problem Set 1.3
**Source:** Main Text
Problems 1–14: Write an equation of the line satisfying the following conditions. Write the equation in the form \(y = mx + b\).
1. It passes through the point \((3, 10)\) and has slope \(= 2\).
Problem 1 Solution
Step 1: We are given slope \(m = 2\) and a point \((3, 10)\). Start with the slope-intercept form and substitute \(m\):
\[y = 2x + b\]
Step 2: Substitute the point \((3, 10)\) into the equation to solve for \(b\):
\[\begin{align} 10 &= 2(3) + b \\ 10 &= 6 + b \\ b &= 4 \end{align}\]
Step 3: Write the complete equation:
\[y = 2x + 4\]
Answer: \(y = 2x + 4\)
Verification: Substituting \((3, 10)\): \(y = 2(3) + 4 = 6 + 4 = 10\) ✓
2. It passes through point \((4, 5)\) and has \(m = 0\).
Problem 2 Solution
Step 1: We are given slope \(m = 0\). A slope of zero means the line is horizontal — the \(y\)-value never changes. Substitute \(m = 0\) into slope-intercept form:
\[y = 0 \cdot x + b = b\]
Step 2: Since the line passes through \((4, 5)\), the constant \(y\)-value must be \(5\):
\[5 = 0(4) + b \implies b = 5\]
Answer: \(y = 5\)
Verification: Substituting \((4, 5)\): \(y = 5\) ✓
3. It passes through \((3, 5)\) and \((2, -1)\).
Problem 3 Solution
Step 1: Calculate the slope using the two points \((3, 5)\) and \((2, -1)\):
\[m = \frac{5 - (-1)}{3 - 2} = \frac{6}{1} = 6\]
Step 2: Substitute \(m = 6\) into slope-intercept form:
\[y = 6x + b\]
Step 3: Substitute the point \((3, 5)\) to solve for \(b\):
\[\begin{align} 5 &= 6(3) + b \\ 5 &= 18 + b \\ b &= -13 \end{align}\]
Step 4: Write the complete equation:
\[y = 6x - 13\]
Answer: \(y = 6x - 13\)
Verification: Substituting \((2, -1)\): \(y = 6(2) - 13 = 12 - 13 = -1\) ✓
4. It has slope \(3\), and its y-intercept equals \(2\).
Problem 4 Solution
Step 1: We are given \(m = 3\) and \(b = 2\) directly. Substitute into slope-intercept form:
\[y = 3x + 2\]
Answer: \(y = 3x + 2\)
Verification: When \(x = 0\): \(y = 3(0) + 2 = 2\) (y-intercept is 2) ✓. The coefficient of \(x\) is 3 (slope is 3) ✓
5. It passes through \((5, -2)\) and \(m = \frac{2}{5}\).
Problem 5 Solution
Step 1: We are given slope \(m = \frac{2}{5}\) and a point \((5, -2)\). Start with slope-intercept form:
\[y = \frac{2}{5}x + b\]
Step 2: Substitute the point \((5, -2)\) to solve for \(b\):
\[\begin{align} -2 &= \frac{2}{5}(5) + b \\ -2 &= 2 + b \\ b &= -4 \end{align}\]
Step 3: Write the complete equation:
\[y = \frac{2}{5}x - 4\]
Answer: \(y = \frac{2}{5}x - 4\)
Verification: Substituting \((5, -2)\): \(y = \frac{2}{5}(5) - 4 = 2 - 4 = -2\) ✓
6. It passes through \((-5, -3)\) and \((10, 0)\).
Problem 6 Solution
Step 1: Calculate the slope using the two points \((-5, -3)\) and \((10, 0)\):
\[m = \frac{0 - (-3)}{10 - (-5)} = \frac{3}{15} = \frac{1}{5}\]
Step 2: Substitute \(m = \frac{1}{5}\) into slope-intercept form:
\[y = \frac{1}{5}x + b\]
Step 3: Substitute the point \((10, 0)\) to solve for \(b\):
\[\begin{align} 0 &= \frac{1}{5}(10) + b \\ 0 &= 2 + b \\ b &= -2 \end{align}\]
Step 4: Write the complete equation:
\[y = \frac{1}{5}x - 2\]
Answer: \(y = \frac{1}{5}x - 2\)
Verification: Substituting \((-5, -3)\): \(y = \frac{1}{5}(-5) - 2 = -1 - 2 = -3\) ✓
7. It passes through \((4, -4)\) and \((5, 3)\).
Problem 7 Solution
Step 1: Calculate the slope using the two points \((4, -4)\) and \((5, 3)\):
\[m = \frac{3 - (-4)}{5 - 4} = \frac{7}{1} = 7\]
Step 2: Substitute \(m = 7\) into slope-intercept form:
\[y = 7x + b\]
Step 3: Substitute the point \((5, 3)\) to solve for \(b\):
\[\begin{align} 3 &= 7(5) + b \\ 3 &= 35 + b \\ b &= -32 \end{align}\]
Step 4: Write the complete equation:
\[y = 7x - 32\]
Answer: \(y = 7x - 32\)
Verification: Substituting \((4, -4)\): \(y = 7(4) - 32 = 28 - 32 = -4\) ✓
8. It passes through \((7, -2)\); its y-intercept is \(5\).
Problem 8 Solution
Step 1: We know \(b = 5\), so the partial equation is:
\[y = mx + 5\]
Step 2: Substitute the point \((7, -2)\) to solve for \(m\):
\[\begin{align} -2 &= m(7) + 5 \\ -2 - 5 &= 7m \\ -7 &= 7m \\ m &= -1 \end{align}\]
Step 3: Write the complete equation:
\[y = -x + 5\]
Answer: \(y = -x + 5\)
Verification: Substituting \((7, -2)\): \(y = -(7) + 5 = -7 + 5 = -2\) ✓
9. It passes through \((2, -5)\) and its x-intercept is \(4\).
Problem 9 Solution
Step 1: An x-intercept of \(4\) means the line passes through the point \((4, 0)\). So we have two points: \((2, -5)\) and \((4, 0)\).
Step 2: Calculate the slope:
\[m = \frac{0 - (-5)}{4 - 2} = \frac{5}{2}\]
Step 3: Substitute \(m = \frac{5}{2}\) into slope-intercept form:
\[y = \frac{5}{2}x + b\]
Step 4: Substitute the point \((4, 0)\) to solve for \(b\):
\[\begin{align} 0 &= \frac{5}{2}(4) + b \\ 0 &= 10 + b \\ b &= -10 \end{align}\]
Step 5: Write the complete equation:
\[y = \frac{5}{2}x - 10\]
Answer: \(y = \frac{5}{2}x - 10\)
Verification: Substituting \((2, -5)\): \(y = \frac{5}{2}(2) - 10 = 5 - 10 = -5\) ✓
10. It's a horizontal line through the point \((2, -1)\).
Problem 10 Solution
Step 1: A horizontal line has slope \(m = 0\). This means \(y\) is the same for every point on the line.
Step 2: Since the line passes through \((2, -1)\), the constant \(y\)-value is \(-1\).
Answer: \(y = -1\)
Verification: The point \((2, -1)\) satisfies \(y = -1\) ✓
11. It passes through \((5, -4)\) and \((1, -4)\).
Problem 11 Solution
Step 1: Calculate the slope using the two points \((5, -4)\) and \((1, -4)\):
\[m = \frac{-4 - (-4)}{1 - 5} = \frac{0}{-4} = 0\]
Step 2: A slope of \(0\) means this is a horizontal line. Both points share the same \(y\)-coordinate of \(-4\).
Answer: \(y = -4\)
Verification: Both \((5, -4)\) and \((1, -4)\) satisfy \(y = -4\) ✓
12. It is a vertical line through the point \((3, -2)\).
Problem 12 Solution
Step 1: A vertical line has an undefined slope and cannot be written in \(y = mx + b\) form. Every point on a vertical line shares the same \(x\)-coordinate.
Step 2: Since the line passes through \((3, -2)\), the constant \(x\)-value is \(3\).
Answer: \(x = 3\)
*Note:* This equation cannot be expressed in slope-intercept form because the slope is undefined.
13. It passes through \((3, -4)\) and \((3, 4)\).
Problem 13 Solution
Step 1: Notice that both points have the same \(x\)-coordinate: \(x = 3\). This means the line is **vertical**.
Step 2: Attempting to compute the slope confirms this:
\[m = \frac{4 - (-4)}{3 - 3} = \frac{8}{0} = \text{undefined}\]
Step 3: Since all points on this line have \(x = 3\), the equation is:
Answer: \(x = 3\)
*Note:* This equation cannot be expressed in slope-intercept form because the slope is undefined.
14. It has x-intercept \(= 3\) and y-intercept \(= 4\).
Problem 14 Solution
Step 1: An x-intercept of \(3\) gives the point \((3, 0)\). A y-intercept of \(4\) gives the point \((0, 4)\).
Step 2: Calculate the slope using these two points:
\[m = \frac{4 - 0}{0 - 3} = \frac{4}{-3} = -\frac{4}{3}\]
Step 3: We already know \(b = 4\) (the y-intercept), so:
\[y = -\frac{4}{3}x + 4\]
Answer: \(y = -\frac{4}{3}x + 4\)
Verification: Substituting \((3, 0)\): \(y = -\frac{4}{3}(3) + 4 = -4 + 4 = 0\) ✓
Problems 15–20: Write an equation of the line satisfying the following conditions. Write the equation in the form \(Ax + By = C\).
15. It passes through \((3, -1)\) and \(m = 2\).
Problem 15 Solution
Step 1: Start with point-slope form using the point \((3, -1)\) and slope \(m = 2\):
\[y - (-1) = 2(x - 3)\]
Step 2: Simplify the left side:
\[y + 1 = 2(x - 3)\]
Step 3: Distribute on the right:
\[y + 1 = 2x - 6\]
Step 4: Rearrange to standard form \(Ax + By = C\) by moving all variable terms to the left and constants to the right:
\[\begin{align} -2x + y &= -6 - 1 \\ -2x + y &= -7 \end{align}\]
Step 5: Multiply by \(-1\) to make the \(x\)-coefficient positive:
\[2x - y = 7\]
Answer: \(2x - y = 7\)
Verification: Substituting \((3, -1)\): \(2(3) - (-1) = 6 + 1 = 7\) ✓
16. It passes through \((-2, 1)\) and \(m = -\frac{3}{2}\).
Problem 16 Solution
Step 1: Start with point-slope form using the point \((-2, 1)\) and slope \(m = -\frac{3}{2}\):
\[y - 1 = -\frac{3}{2}(x - (-2))\]
Step 2: Simplify:
\[y - 1 = -\frac{3}{2}(x + 2)\]
Step 3: Multiply both sides by \(2\) to eliminate the fraction:
\[2(y - 1) = -3(x + 2)\]
Step 4: Distribute:
\[2y - 2 = -3x - 6\]
Step 5: Rearrange to standard form:
\[\begin{align} 3x + 2y &= -6 + 2 \\ 3x + 2y &= -4 \end{align}\]
Answer: \(3x + 2y = -4\)
Verification: Substituting \((-2, 1)\): \(3(-2) + 2(1) = -6 + 2 = -4\) ✓
17. It passes through \((-4, -2)\) and \(m = \frac{3}{4}\).
Problem 17 Solution
Step 1: Start with point-slope form using the point \((-4, -2)\) and slope \(m = \frac{3}{4}\):
\[y - (-2) = \frac{3}{4}(x - (-4))\]
Step 2: Simplify:
\[y + 2 = \frac{3}{4}(x + 4)\]
Step 3: Multiply both sides by \(4\) to eliminate the fraction:
\[4(y + 2) = 3(x + 4)\]
Step 4: Distribute:
\[4y + 8 = 3x + 12\]
Step 5: Rearrange to standard form:
\[\begin{align} -3x + 4y &= 12 - 8 \\ -3x + 4y &= 4 \end{align}\]
Step 6: Multiply by \(-1\) to make the \(x\)-coefficient positive:
\[3x - 4y = -4\]
Answer: \(3x - 4y = -4\)
Verification: Substituting \((-4, -2)\): \(3(-4) - 4(-2) = -12 + 8 = -4\) ✓
18. Its x-intercept equals \(3\), and \(m = -\frac{5}{3}\).
Problem 18 Solution
Step 1: An x-intercept of \(3\) means the line passes through the point \((3, 0)\). Use point-slope form with \((3, 0)\) and \(m = -\frac{5}{3}\):
\[y - 0 = -\frac{5}{3}(x - 3)\]
Step 2: Multiply both sides by \(3\) to eliminate the fraction:
\[3y = -5(x - 3)\]
Step 3: Distribute:
\[3y = -5x + 15\]
Step 4: Rearrange to standard form:
\[5x + 3y = 15\]
Answer: \(5x + 3y = 15\)
Verification: Substituting \((3, 0)\): \(5(3) + 3(0) = 15 + 0 = 15\) ✓
19. It passes through \((2, -3)\) and \((5, 1)\).
Problem 19 Solution
Step 1: Calculate the slope using the two points \((2, -3)\) and \((5, 1)\):
\[m = \frac{1 - (-3)}{5 - 2} = \frac{4}{3}\]
Step 2: Use point-slope form with the point \((5, 1)\):
\[y - 1 = \frac{4}{3}(x - 5)\]
Step 3: Multiply both sides by \(3\) to eliminate the fraction:
\[3(y - 1) = 4(x - 5)\]
Step 4: Distribute:
\[3y - 3 = 4x - 20\]
Step 5: Rearrange to standard form:
\[\begin{align} -4x + 3y &= -20 + 3 \\ -4x + 3y &= -17 \end{align}\]
Step 6: Multiply by \(-1\):
\[4x - 3y = 17\]
Answer: \(4x - 3y = 17\)
Verification: Substituting \((2, -3)\): \(4(2) - 3(-3) = 8 + 9 = 17\) ✓. Substituting \((5, 1)\): \(4(5) - 3(1) = 20 - 3 = 17\) ✓
20. It passes through \((1, -3)\) and \((-5, 5)\).
Problem 20 Solution
Step 1: Calculate the slope using the two points \((1, -3)\) and \((-5, 5)\):
\[m = \frac{5 - (-3)}{-5 - 1} = \frac{8}{-6} = -\frac{4}{3}\]
Step 2: Use point-slope form with the point \((1, -3)\):
\[y - (-3) = -\frac{4}{3}(x - 1)\]
Step 3: Simplify and multiply both sides by \(3\):
\[3(y + 3) = -4(x - 1)\]
Step 4: Distribute:
\[3y + 9 = -4x + 4\]
Step 5: Rearrange to standard form:
\[\begin{align} 4x + 3y &= 4 - 9 \\ 4x + 3y &= -5 \end{align}\]
Answer: \(4x + 3y = -5\)
Verification: Substituting \((1, -3)\): \(4(1) + 3(-3) = 4 - 9 = -5\) ✓. Substituting \((-5, 5)\): \(4(-5) + 3(5) = -20 + 15 = -5\) ✓
Problems 21–26: Write an equation of the line satisfying the following conditions. Write the equation in point-slope form \(y - y_{1} = m(x - x_{1})\).
21. It passes through \((2, -3)\) and \((5, 1)\).
Problem 21 Solution
Step 1: Calculate the slope using the two points \((2, -3)\) and \((5, 1)\):
\[m = \frac{1 - (-3)}{5 - 2} = \frac{4}{3}\]
Step 2: Substitute into point-slope form using the point \((2, -3)\):
\[y - (-3) = \frac{4}{3}(x - 2)\]
Step 3: Simplify the left side:
\[y + 3 = \frac{4}{3}(x - 2)\]
Answer: \(y + 3 = \frac{4}{3}(x - 2)\)
*Note:* Using the other point \((5, 1)\) gives the equivalent form \(y - 1 = \frac{4}{3}(x - 5)\).
Verification: Substituting \((5, 1)\): LHS \(= 1 + 3 = 4\); RHS \(= \frac{4}{3}(5 - 2) = \frac{4}{3}(3) = 4\) ✓
22. It passes through \((1, -3)\) and \((-5, 2)\).
Problem 22 Solution
Step 1: Calculate the slope using the two points \((1, -3)\) and \((-5, 2)\):
\[m = \frac{2 - (-3)}{-5 - 1} = \frac{5}{-6} = -\frac{5}{6}\]
Step 2: Substitute into point-slope form using the point \((1, -3)\):
\[y - (-3) = -\frac{5}{6}(x - 1)\]
Step 3: Simplify:
\[y + 3 = -\frac{5}{6}(x - 1)\]
Answer: \(y + 3 = -\frac{5}{6}(x - 1)\)
*Note:* Using the other point \((-5, 2)\) gives the equivalent form \(y - 2 = -\frac{5}{6}(x + 5)\).
Verification: Substituting \((-5, 2)\): LHS \(= 2 + 3 = 5\); RHS \(= -\frac{5}{6}(-5 - 1) = -\frac{5}{6}(-6) = 5\) ✓
23. It passes through \((6, -2)\) and \((0, 2)\).
Problem 23 Solution
Step 1: Calculate the slope using the two points \((6, -2)\) and \((0, 2)\):
\[m = \frac{2 - (-2)}{0 - 6} = \frac{4}{-6} = -\frac{2}{3}\]
Step 2: Substitute into point-slope form using the point \((6, -2)\):
\[y - (-2) = -\frac{2}{3}(x - 6)\]
Step 3: Simplify:
\[y + 2 = -\frac{2}{3}(x - 6)\]
Answer: \(y + 2 = -\frac{2}{3}(x - 6)\)
*Note:* Using the other point \((0, 2)\) gives the equivalent form \(y - 2 = -\frac{2}{3}(x - 0)\) or \(y - 2 = -\frac{2}{3}x\).
Verification: Substituting \((0, 2)\): LHS \(= 2 + 2 = 4\); RHS \(= -\frac{2}{3}(0 - 6) = -\frac{2}{3}(-6) = 4\) ✓
24. It passes through \((8, 2)\) and \((-7, -4)\).
Problem 24 Solution
Step 1: Calculate the slope using the two points \((8, 2)\) and \((-7, -4)\):
\[m = \frac{-4 - 2}{-7 - 8} = \frac{-6}{-15} = \frac{2}{5}\]
Step 2: Substitute into point-slope form using the point \((8, 2)\):
\[y - 2 = \frac{2}{5}(x - 8)\]
Answer: \(y - 2 = \frac{2}{5}(x - 8)\)
*Note:* Using the other point \((-7, -4)\) gives the equivalent form \(y + 4 = \frac{2}{5}(x + 7)\).
Verification: Substituting \((-7, -4)\): LHS \(= -4 - 2 = -6\); RHS \(= \frac{2}{5}(-7 - 8) = \frac{2}{5}(-15) = -6\) ✓
25. It passes through \((-12, 7)\) and has slope \(= -\frac{1}{3}\).
Problem 25 Solution
Step 1: We are given the point \((-12, 7)\) and slope \(m = -\frac{1}{3}\). Substitute directly into point-slope form:
\[y - 7 = -\frac{1}{3}(x - (-12))\]
Step 2: Simplify the expression inside the parentheses:
\[y - 7 = -\frac{1}{3}(x + 12)\]
Answer: \(y - 7 = -\frac{1}{3}(x + 12)\)
Verification: Substituting \((-12, 7)\): LHS \(= 7 - 7 = 0\); RHS \(= -\frac{1}{3}(-12 + 12) = -\frac{1}{3}(0) = 0\) ✓
26. It passes through \((8, -7)\) and has slope \(\frac{3}{4}\).
Problem 26 Solution
Step 1: We are given the point \((8, -7)\) and slope \(m = \frac{3}{4}\). Substitute directly into point-slope form:
\[y - (-7) = \frac{3}{4}(x - 8)\]
Step 2: Simplify the left side:
\[y + 7 = \frac{3}{4}(x - 8)\]
Answer: \(y + 7 = \frac{3}{4}(x - 8)\)
Verification: Substituting \((8, -7)\): LHS \(= -7 + 7 = 0\); RHS \(= \frac{3}{4}(8 - 8) = \frac{3}{4}(0) = 0\) ✓