Answer :
To solve this problem, we need to determine the equation of a line that passes through a given point and has a specified slope in three different forms: point-slope form, slope-intercept form, and general form.
Given:
- Point: (1, 2)
- Slope: 3
### (a) Point-Slope Form
The point-slope form of the equation of a line is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\( (x_1, y_1) \)[/tex] is the given point and [tex]\( m \)[/tex] is the slope.
Plugging in the given point (1, 2) and the slope 3:
[tex]\[ y - 2 = 3(x - 1) \][/tex]
So, the point-slope form of the equation is:
[tex]\[ y - 2 = 3(x - 1) \][/tex]
### (b) Slope-Intercept Form
The slope-intercept form of the equation of a line is given by:
[tex]\[ y = mx + b \][/tex]
where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept.
Using the point (1, 2) and the slope 3, we first need to determine [tex]\( b \)[/tex].
We know:
[tex]\[ y = 3x + b \][/tex]
At the point (1, 2):
[tex]\[ 2 = 3(1) + b \][/tex]
[tex]\[ 2 = 3 + b \][/tex]
[tex]\[ b = 2 - 3 \][/tex]
[tex]\[ b = -1 \][/tex]
So, the slope-intercept form of the equation is:
[tex]\[ y = 3x - 1 \][/tex]
### (c) General Form
The general form of the equation of a line is given by:
[tex]\[ Ax + By + C = 0 \][/tex]
where [tex]\( A, B, \)[/tex] and [tex]\( C \)[/tex] are integers, and [tex]\( A > 0 \)[/tex].
Starting from the slope-intercept form [tex]\( y = 3x - 1 \)[/tex], we need to rearrange it into the general form.
[tex]\[ y = 3x - 1 \][/tex]
Subtract [tex]\( y \)[/tex] from both sides:
[tex]\[ 0 = 3x - y - 1 \][/tex]
Rewriting it for clarity:
[tex]\[ 3x - y - 1 = 0 \][/tex]
So, the general form of the equation is:
[tex]\[ 3x - y - 1 = 0 \][/tex]
In summary, the equations in the different forms are:
- Point-Slope Form: [tex]\( y - 2 = 3(x - 1) \)[/tex]
- Slope-Intercept Form: [tex]\( y = 3x - 1 \)[/tex]
- General Form: [tex]\( 3x - y - 1 = 0 \)[/tex]
Given:
- Point: (1, 2)
- Slope: 3
### (a) Point-Slope Form
The point-slope form of the equation of a line is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\( (x_1, y_1) \)[/tex] is the given point and [tex]\( m \)[/tex] is the slope.
Plugging in the given point (1, 2) and the slope 3:
[tex]\[ y - 2 = 3(x - 1) \][/tex]
So, the point-slope form of the equation is:
[tex]\[ y - 2 = 3(x - 1) \][/tex]
### (b) Slope-Intercept Form
The slope-intercept form of the equation of a line is given by:
[tex]\[ y = mx + b \][/tex]
where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept.
Using the point (1, 2) and the slope 3, we first need to determine [tex]\( b \)[/tex].
We know:
[tex]\[ y = 3x + b \][/tex]
At the point (1, 2):
[tex]\[ 2 = 3(1) + b \][/tex]
[tex]\[ 2 = 3 + b \][/tex]
[tex]\[ b = 2 - 3 \][/tex]
[tex]\[ b = -1 \][/tex]
So, the slope-intercept form of the equation is:
[tex]\[ y = 3x - 1 \][/tex]
### (c) General Form
The general form of the equation of a line is given by:
[tex]\[ Ax + By + C = 0 \][/tex]
where [tex]\( A, B, \)[/tex] and [tex]\( C \)[/tex] are integers, and [tex]\( A > 0 \)[/tex].
Starting from the slope-intercept form [tex]\( y = 3x - 1 \)[/tex], we need to rearrange it into the general form.
[tex]\[ y = 3x - 1 \][/tex]
Subtract [tex]\( y \)[/tex] from both sides:
[tex]\[ 0 = 3x - y - 1 \][/tex]
Rewriting it for clarity:
[tex]\[ 3x - y - 1 = 0 \][/tex]
So, the general form of the equation is:
[tex]\[ 3x - y - 1 = 0 \][/tex]
In summary, the equations in the different forms are:
- Point-Slope Form: [tex]\( y - 2 = 3(x - 1) \)[/tex]
- Slope-Intercept Form: [tex]\( y = 3x - 1 \)[/tex]
- General Form: [tex]\( 3x - y - 1 = 0 \)[/tex]