Respuesta :
Answer:
1) The equation of the line in slope-intercept form is [tex]y = 5\cdot x +9[/tex]. The equation of the line in standard form is [tex]-5\cdot x + y = 9[/tex].
2) The equation of the line in slope-intercept form is [tex]y = \frac{2}{5}\cdot x +\frac{14}{5}[/tex]. The equation of the line in standard form is [tex]-2\cdot x +5\cdot y = 14[/tex].
3) The equation of the line in slope-intercept form is [tex]y = 3\cdot x +4[/tex]. The equation of the line in standard form is [tex]-3\cdot x +y = 4[/tex].
4) The equation of the line in slope-intercept form is [tex]y = 2\cdot x + 6[/tex]. The equation of the line in standard form is [tex]-2\cdot x +y = 6[/tex].
5) The equation of the line in slope-intercept form is [tex]y = \frac{5}{6}\cdot x -\frac{7}{6}[/tex]. The equation of the line in standard from is [tex]-5\cdot x + 6\cdot y = -7[/tex].
Step-by-step explanation:
1) We begin with the slope-intercept form and substitute all known values and calculate the y-intercept: ([tex]m = 5[/tex], [tex]x = -1[/tex], [tex]y = 4[/tex])
[tex]4 = (5)\cdot (-1)+b[/tex]
[tex]4 = -5 +b[/tex]
[tex]b = 9[/tex]
The equation of the line in slope-intercept form is [tex]y = 5\cdot x +9[/tex].
Then, we obtain the standard form by algebraic handling:
[tex]-5\cdot x + y = 9[/tex]
The equation of the line in standard form is [tex]-5\cdot x + y = 9[/tex].
2) We begin with a system of linear equations based on the slope-intercept form: ([tex]x_{1} = 3[/tex], [tex]y_{1} = 4[/tex], [tex]x_{2} = -2[/tex], [tex]y_{2} = 2[/tex])
[tex]3\cdot m + b = 4[/tex] (Eq. 1)
[tex]-2\cdot m + b = 2[/tex] (Eq. 2)
From (Eq. 1), we find that:
[tex]b = 4-3\cdot m[/tex]
And by substituting on (Eq. 2), we conclude that slope of the equation of the line is:
[tex]-2\cdot m +4-3\cdot m = 2[/tex]
[tex]-5\cdot m = -2[/tex]
[tex]m = \frac{2}{5}[/tex]
And from (Eq. 1) we find that the y-Intercept is:
[tex]b=4-3\cdot \left(\frac{2}{5} \right)[/tex]
[tex]b = 4-\frac{6}{5}[/tex]
[tex]b = \frac{14}{5}[/tex]
The equation of the line in slope-intercept form is [tex]y = \frac{2}{5}\cdot x +\frac{14}{5}[/tex].
Then, we obtain the standard form by algebraic handling:
[tex]-\frac{2}{5}\cdot x +y = \frac{14}{5}[/tex]
[tex]-2\cdot x +5\cdot y = 14[/tex]
The equation of the line in standard form is [tex]-2\cdot x +5\cdot y = 14[/tex].
3) By using the slope-intercept form, we obtain the equation of the line by direct substitution: ([tex]m = 3[/tex], [tex]b = 4[/tex])
[tex]y = 3\cdot x +4[/tex]
The equation of the line in slope-intercept form is [tex]y = 3\cdot x +4[/tex].
Then, we obtain the standard form by algebraic handling:
[tex]-3\cdot x +y = 4[/tex]
The equation of the line in standard form is [tex]-3\cdot x +y = 4[/tex].
4) We begin with a system of linear equations based on the slope-intercept form: ([tex]x_{1} = -3[/tex], [tex]y_{1} = 0[/tex], [tex]x_{2} = 0[/tex], [tex]y_{2} = 6[/tex])
[tex]-3\cdot m + b = 0[/tex] (Eq. 3)
[tex]b = 6[/tex] (Eq. 4)
By applying (Eq. 4) on (Eq. 3), we find that the slope of the equation of the line is:
[tex]-3\cdot m+6 = 0[/tex]
[tex]3\cdot m = 6[/tex]
[tex]m = 2[/tex]
The equation of the line in slope-intercept form is [tex]y = 2\cdot x + 6[/tex].
Then, we obtain the standard form by algebraic handling:
[tex]-2\cdot x +y = 6[/tex]
The equation of the line in standard form is [tex]-2\cdot x +y = 6[/tex].
5) We begin with a system of linear equations based on the slope-intercept form: ([tex]x_{1} = -1[/tex], [tex]y_{1} = -2[/tex], [tex]x_{2} = 5[/tex], [tex]y_{2} = 3[/tex])
[tex]-m+b = -2[/tex] (Eq. 5)
[tex]5\cdot m +b = 3[/tex] (Eq. 6)
From (Eq. 5), we find that:
[tex]b = -2+m[/tex]
And by substituting on (Eq. 6), we conclude that slope of the equation of the line is:
[tex]5\cdot m -2+m = 3[/tex]
[tex]6\cdot m = 5[/tex]
[tex]m = \frac{5}{6}[/tex]
And from (Eq. 5) we find that the y-Intercept is:
[tex]b = -2+\frac{5}{6}[/tex]
[tex]b = -\frac{7}{6}[/tex]
The equation of the line in slope-intercept form is [tex]y = \frac{5}{6}\cdot x -\frac{7}{6}[/tex].
Then, we obtain the standard form by algebraic handling:
[tex]-\frac{5}{6}\cdot x +y =-\frac{7}{6}[/tex]
[tex]-5\cdot x + 6\cdot y = -7[/tex]
The equation of the line in standard from is [tex]-5\cdot x + 6\cdot y = -7[/tex].
A linear equation can be written in slope intercept and in standard form.
(1) Slope = 5, Point = (-1,4)
The equation is calculated as:
[tex]\mathbf{y = m(x - x_1) + y_1}[/tex]
So, we have:
[tex]\mathbf{y = 5(x + 1) + 4}[/tex]
[tex]\mathbf{y = 5x + 5 + 4}[/tex]
[tex]\mathbf{y = 5x + 9}[/tex]
Subtract 5x from both sides
[tex]\mathbf{-5x + y = 9}[/tex]
So, the equations are:
- Slope intercept: [tex]\mathbf{y = 5x + 9}[/tex]
- Standard form: [tex]\mathbf{-5x + y = 9}[/tex].
(2) Points = (3,4) and (-2,2)
First, we calculate the slope (m)
[tex]\mathbf{m =\frac{y_2 - y_1}{x_2 - x_1}}[/tex]
So, we have:
[tex]\mathbf{m =\frac{2-4}{-2-3} = \frac{2}{5}}[/tex]
The equation is calculated as:
[tex]\mathbf{y = m(x - x_1) + y_1}[/tex]
So, we have:
[tex]\mathbf{y = \frac 25(x - 3) + 4}[/tex]
[tex]\mathbf{y = \frac 25x - \frac{6}{5} + 4}[/tex]
[tex]\mathbf{y = \frac 25x + \frac{-6 + 20}{5}}[/tex]
[tex]\mathbf{y = \frac 25x - \frac{14}{5}}[/tex]
Multiply through by 5
[tex]\mathbf{5y = 2x - 14}[/tex]
Rewrite as:
[tex]\mathbf{2x - 5y = 14}[/tex]
So, the equations are:
- Slope intercept: [tex]\mathbf{y = \frac 25x - \frac{14}{5}}[/tex]
- Standard form: [tex]\mathbf{2x - 5y = 14}[/tex].
(3) Slope = 3, y-intercept = 4
The m and b in y = mx + b represent slope and y-intercept, respectively.
So, we have:
[tex]\mathbf{y = 3x + 4}[/tex]
Rewrite as:
[tex]\mathbf{3x - y = -4}[/tex]
So, the equations are:
- Slope intercept: [tex]\mathbf{y = 3x + 4}[/tex]
- Standard form: [tex]\mathbf{3x - y = -4}[/tex].
(4) x-intercept = -3, y-intercept = 6
Calculate the slope using:
[tex]\mathbf{m = -\frac{y-intercept}{x-intercept}}[/tex]
[tex]\mathbf{m = -\frac{6}{-3}}[/tex]
[tex]\mathbf{m = 2}[/tex]
The m and b in y = mx + b represent slope and y-intercept, respectively.
So, we have:
[tex]\mathbf{y = 2x + 6}[/tex]
Rewrite as:
[tex]\mathbf{2x - y =-6}[/tex]
So, the equations are:
- Slope intercept: [tex]\mathbf{y = 2x + 6}[/tex]
- Standard form: [tex]\mathbf{2x - y =-6}[/tex].
(5) Points = (-1,-2) and (5,3)
First, we calculate the slope (m)
[tex]\mathbf{m =\frac{y_2 - y_1}{x_2 - x_1}}[/tex]
So, we have:
[tex]\mathbf{m =\frac{3--2}{5--1} = \frac{5}{6}}[/tex]
The equation is calculated as:
[tex]\mathbf{y = m(x - x_1) + y_1}[/tex]
So, we have:
[tex]\mathbf{y = \frac 56(x + 1) - 2}[/tex]
[tex]\mathbf{y = \frac 56x + \frac 56 - 2}[/tex]
[tex]\mathbf{y = \frac 56x + \frac{5 - 12}{6}}[/tex]
[tex]\mathbf{y = \frac 56x - \frac{ 7}{6}}[/tex]
Multiply through by 6
[tex]\mathbf{6y = 5x - 7}[/tex]
Rewrite as:
[tex]\mathbf{5x - 6y =7}[/tex]
So, the equations are:
- Slope intercept: [tex]\mathbf{y = \frac 56x - \frac{ 7}{6}}[/tex]
- Standard form: [tex]\mathbf{5x - 6y =7}[/tex]
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