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Find the equation of a line glven the point and slope below. Arrange your answer in the form y = mx + b, where b is the

Greeley [361]

Answer:

b=-79

Step-by-step explanation:

(10,11) means x=10,y=11

m=9 b=?

y=mx+b

Make b the subject of formula

b=y-mx

b=11-9(10)

b=11-90

b=-79

4 0

3 years ago

Find the equation of the line . Write in slope intercept form and in standard form. (SHOW YOUR SOLUTION)

AlladinOne [14]

Answer:

1) The slope-intercept and standard forms are $y = -5\cdot x + 1$ and $5\cdot x +y = 1$ , respectively.

2) The slope-intercept form of the line is $y = \frac{5}{2}\cdot x -\frac{9}{2}$ . The standard form of the line is $-5\cdot x +2\cdot y = -9$ .

3) The slope-intercept form of the line is $y = \frac{5}{2}\cdot x + 5$ . The standard form of the line is $-5\cdot x +2\cdot y = 10$ .

4) The slope-intercept and standard forms of the family of lines are $y = \frac{2}{7}\cdot x -\frac{c}{7}$ and $2\cdot x -7\cdot y = c$ , $\forall \,c \in \mathbb{R}$ , respectively.

5) The slope-intercept form of the line is $y = 2\cdot x-7$ . The standard form of the line is $-2\cdot x +y = -7$ .

Step-by-step explanation:

From Analytical Geometry we know that the slope-intercept form of the line is represented by:

$y = m\cdot x + b$ (1)

Where:

$x$ - Independent variable, dimensionless.

$m$ - Slope, dimensionless.

$b$ - y-Intercept, dimensionless.

$y$ - Dependent variable, dimensionless.

In addition, the standard form of the line is represented by the following model:

$a\cdot x + b \cdot y = c$ (2)

Where $a$ , $b$ are constant coefficients, dimensionless.

Now we process to resolve each problem:

1) If we know that $m = -5$ and $b = 1$ , then we know that the slope-intercept form of the line is:

$y = -5\cdot x + 1$ (3)

And the standard form is found after some algebraic handling:

$5\cdot x +y = 1$ (4)

The slope-intercept and standard forms are $y = -5\cdot x + 1$ and $5\cdot x +y = 1$ , respectively.

2) From Geometry we know that a line can be formed by two distinct points on a plane. If we know that $(x_{1},y_{1})=(1,-2)$ and $(x_{2},y_{2}) = (3,3)$ , then we construct the following system of linear equations:

$m+b= -2$ (5)

$3\cdot m +b = 3$ (6)

The solution of the system is:

$m = \frac{5}{2}$ , $b = -\frac{9}{2}$

The slope-intercept form of the line is $y = \frac{5}{2}\cdot x -\frac{9}{2}$ .

And the standard form is found after some algebraic handling:

$-\frac{5}{2}\cdot x +y = -\frac{9}{2}$

$-5\cdot x +2\cdot y = -9$ (7)

The standard form of the line is $-5\cdot x +2\cdot y = -9$ .

3) From Geometry we know that a line can be formed by two distinct points on a plane. If we know that $(x_{1},y_{1})=(-2,0)$ and $(x_{2},y_{2}) = (0,5)$ , then we construct the following system of linear equations:

$-2\cdot m +b = 0$ (8)

$b = 5$ (9)

The solution of the system is:

$m =\frac{5}{2}$ , $b = 5$

The slope-intercept form of the line is $y = \frac{5}{2}\cdot x + 5$ .

And the standard form is found after some algebraic handling:

$-\frac{5}{2}\cdot x+y =5$

$-5\cdot x +2\cdot y = 10$ (10)

The standard form of the line is $-5\cdot x +2\cdot y = 10$ .

4) If we know that $a = 2$ and $b = -7$ , then the standard form of the family of lines is:

$2\cdot x -7\cdot y = c$ , $\forall \,c \in \mathbb{R}$

And the standard form is found after some algebraic handling:

$-7\cdot y = -2\cdot x +c$

$y = \frac{2}{7}\cdot x -\frac{c}{7}$ , $\forall \,c\in\mathbb{R}$ (11)

The slope-intercept and standard forms of the family of lines are $y = \frac{2}{7}\cdot x -\frac{c}{7}$ and $2\cdot x -7\cdot y = c$ , $\forall \,c \in \mathbb{R}$ , respectively.

5) If we know that $(x,y) = (3,-1)$ and $m = 2$ , then the y-intercept of the line is: