Respuesta :
Answer:
1) The slope-intercept and standard forms are [tex]y = -5\cdot x + 1[/tex] and [tex]5\cdot x +y = 1[/tex], respectively.
2) The slope-intercept form of the line is [tex]y = \frac{5}{2}\cdot x -\frac{9}{2}[/tex]. The standard form of the line is [tex]-5\cdot x +2\cdot y = -9[/tex].
3) The slope-intercept form of the line is [tex]y = \frac{5}{2}\cdot x + 5[/tex]. The standard form of the line is [tex]-5\cdot x +2\cdot y = 10[/tex].
4) The slope-intercept and standard forms of the family of lines are [tex]y = \frac{2}{7}\cdot x -\frac{c}{7}[/tex] and [tex]2\cdot x -7\cdot y = c[/tex], [tex]\forall \,c \in \mathbb{R}[/tex], respectively.
5) The slope-intercept form of the line is [tex]y = 2\cdot x-7[/tex]. The standard form of the line is [tex]-2\cdot x +y = -7[/tex].
Step-by-step explanation:
From Analytical Geometry we know that the slope-intercept form of the line is represented by:
[tex]y = m\cdot x + b[/tex] (1)
Where:
[tex]x[/tex] - Independent variable, dimensionless.
[tex]m[/tex] - Slope, dimensionless.
[tex]b[/tex] - y-Intercept, dimensionless.
[tex]y[/tex] - Dependent variable, dimensionless.
In addition, the standard form of the line is represented by the following model:
[tex]a\cdot x + b \cdot y = c[/tex] (2)
Where [tex]a[/tex], [tex]b[/tex] are constant coefficients, dimensionless.
Now we process to resolve each problem:
1) If we know that [tex]m = -5[/tex] and [tex]b = 1[/tex], then we know that the slope-intercept form of the line is:
[tex]y = -5\cdot x + 1[/tex] (3)
And the standard form is found after some algebraic handling:
[tex]5\cdot x +y = 1[/tex] (4)
The slope-intercept and standard forms are [tex]y = -5\cdot x + 1[/tex] and [tex]5\cdot x +y = 1[/tex], respectively.
2) From Geometry we know that a line can be formed by two distinct points on a plane. If we know that [tex](x_{1},y_{1})=(1,-2)[/tex] and [tex](x_{2},y_{2}) = (3,3)[/tex], then we construct the following system of linear equations:
[tex]m+b= -2[/tex] (5)
[tex]3\cdot m +b = 3[/tex] (6)
The solution of the system is:
[tex]m = \frac{5}{2}[/tex], [tex]b = -\frac{9}{2}[/tex]
The slope-intercept form of the line is [tex]y = \frac{5}{2}\cdot x -\frac{9}{2}[/tex].
And the standard form is found after some algebraic handling:
[tex]-\frac{5}{2}\cdot x +y = -\frac{9}{2}[/tex]
[tex]-5\cdot x +2\cdot y = -9[/tex] (7)
The standard form of the line is [tex]-5\cdot x +2\cdot y = -9[/tex].
3) From Geometry we know that a line can be formed by two distinct points on a plane. If we know that [tex](x_{1},y_{1})=(-2,0)[/tex] and [tex](x_{2},y_{2}) = (0,5)[/tex], then we construct the following system of linear equations:
[tex]-2\cdot m +b = 0[/tex] (8)
[tex]b = 5[/tex] (9)
The solution of the system is:
[tex]m =\frac{5}{2}[/tex], [tex]b = 5[/tex]
The slope-intercept form of the line is [tex]y = \frac{5}{2}\cdot x + 5[/tex].
And the standard form is found after some algebraic handling:
[tex]-\frac{5}{2}\cdot x+y =5[/tex]
[tex]-5\cdot x +2\cdot y = 10[/tex] (10)
The standard form of the line is [tex]-5\cdot x +2\cdot y = 10[/tex].
4) If we know that [tex]a = 2[/tex] and [tex]b = -7[/tex], then the standard form of the family of lines is:
[tex]2\cdot x -7\cdot y = c[/tex], [tex]\forall \,c \in \mathbb{R}[/tex]
And the standard form is found after some algebraic handling:
[tex]-7\cdot y = -2\cdot x +c[/tex]
[tex]y = \frac{2}{7}\cdot x -\frac{c}{7}[/tex], [tex]\forall \,c\in\mathbb{R}[/tex] (11)
The slope-intercept and standard forms of the family of lines are [tex]y = \frac{2}{7}\cdot x -\frac{c}{7}[/tex] and [tex]2\cdot x -7\cdot y = c[/tex], [tex]\forall \,c \in \mathbb{R}[/tex], respectively.
5) If we know that [tex](x,y) = (3,-1)[/tex] and [tex]m = 2[/tex], then the y-intercept of the line is:
[tex]3\cdot 2 + b = -1[/tex]
[tex]b = -7[/tex]
Then, the slope-intercept form of the line is [tex]y = 2\cdot x-7[/tex].
And the standard form is found after some algebraic handling:
[tex]-2\cdot x +y = -7[/tex] (12)
The standard form of the line is [tex]-2\cdot x +y = -7[/tex].
