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Sure! To graph a line with a given slope and a point that it passes through, follow these steps:
### Step 1: Understand the Slope-Intercept Form
The slope-intercept form of a line is given by the equation:
[tex]\[ y = mx + b \][/tex]
where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept.
### Step 2: Plug in the Slope and Point to Find the Y-Intercept
Given:
- Slope ([tex]\( m \)[/tex]) = 1
- Point ([tex]\( x_1, y_1 \)[/tex]) = (-3, -3)
We will substitute the point into the slope-intercept equation to solve for [tex]\( b \)[/tex]:
[tex]\[ y_1 = mx_1 + b \][/tex]
Substitute the given values:
[tex]\[ -3 = 1(-3) + b \][/tex]
Simplify and solve for [tex]\( b \)[/tex]:
[tex]\[ -3 = -3 + b \][/tex]
[tex]\[ b = 0 \][/tex]
### Step 3: Write the Equation of the Line
Now that we have [tex]\( m = 1 \)[/tex] and [tex]\( b = 0 \)[/tex], the equation of the line is:
[tex]\[ y = x \][/tex]
### Step 4: Generate Points to Plot the Line
To plot this line, we can generate and plot several points that satisfy the equation [tex]\( y = x \)[/tex].
For example:
- If [tex]\( x = -5 \)[/tex], then [tex]\( y = -5 \)[/tex]
- If [tex]\( x = -3 \)[/tex], then [tex]\( y = -3 \)[/tex]
- If [tex]\( x = 0 \)[/tex], then [tex]\( y = 0 \)[/tex]
- If [tex]\( x = 3 \)[/tex], then [tex]\( y = 3 \)[/tex]
- If [tex]\( x = 5 \)[/tex], then [tex]\( y = 5 \)[/tex]
### Step 5: Plot the Line and the Given Point
Using the points generated, plot the line on a coordinate plane, and add the given point (-3, -3) on the graph.
Here's how it will look:
1. Draw the x-axis and y-axis.
2. Plot the points (-5, -5), (-3, -3), (0, 0), (3, 3), and (5, 5).
3. Draw a straight line through these points.
4. Highlight the point (-3, -3) by marking it on the graph.
The resulting graph is a line passing through the points listed, and it includes the specific point (-3, -3), verifying that the line [tex]\( y = x \)[/tex] correctly passes through (-3, -3).
The graphical representation should show a line with a slope of 1 (meaning it rises one unit vertically for every one unit it moves horizontally) that crosses the origin (0,0) and passes through all the points where [tex]\( y = x \)[/tex].
### Step 1: Understand the Slope-Intercept Form
The slope-intercept form of a line is given by the equation:
[tex]\[ y = mx + b \][/tex]
where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept.
### Step 2: Plug in the Slope and Point to Find the Y-Intercept
Given:
- Slope ([tex]\( m \)[/tex]) = 1
- Point ([tex]\( x_1, y_1 \)[/tex]) = (-3, -3)
We will substitute the point into the slope-intercept equation to solve for [tex]\( b \)[/tex]:
[tex]\[ y_1 = mx_1 + b \][/tex]
Substitute the given values:
[tex]\[ -3 = 1(-3) + b \][/tex]
Simplify and solve for [tex]\( b \)[/tex]:
[tex]\[ -3 = -3 + b \][/tex]
[tex]\[ b = 0 \][/tex]
### Step 3: Write the Equation of the Line
Now that we have [tex]\( m = 1 \)[/tex] and [tex]\( b = 0 \)[/tex], the equation of the line is:
[tex]\[ y = x \][/tex]
### Step 4: Generate Points to Plot the Line
To plot this line, we can generate and plot several points that satisfy the equation [tex]\( y = x \)[/tex].
For example:
- If [tex]\( x = -5 \)[/tex], then [tex]\( y = -5 \)[/tex]
- If [tex]\( x = -3 \)[/tex], then [tex]\( y = -3 \)[/tex]
- If [tex]\( x = 0 \)[/tex], then [tex]\( y = 0 \)[/tex]
- If [tex]\( x = 3 \)[/tex], then [tex]\( y = 3 \)[/tex]
- If [tex]\( x = 5 \)[/tex], then [tex]\( y = 5 \)[/tex]
### Step 5: Plot the Line and the Given Point
Using the points generated, plot the line on a coordinate plane, and add the given point (-3, -3) on the graph.
Here's how it will look:
1. Draw the x-axis and y-axis.
2. Plot the points (-5, -5), (-3, -3), (0, 0), (3, 3), and (5, 5).
3. Draw a straight line through these points.
4. Highlight the point (-3, -3) by marking it on the graph.
The resulting graph is a line passing through the points listed, and it includes the specific point (-3, -3), verifying that the line [tex]\( y = x \)[/tex] correctly passes through (-3, -3).
The graphical representation should show a line with a slope of 1 (meaning it rises one unit vertically for every one unit it moves horizontally) that crosses the origin (0,0) and passes through all the points where [tex]\( y = x \)[/tex].
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