Mathematics, 07.11.2020 14:00 pedro48

Learning Task 3. Find the equation of the line. Do it in your notebook. (a) In slope intercept form y = mx + b
(b) in standard form ax + by = 0
1. The slope is 5 passing through (-1,4).
2. The line passes through point (3, 4) and (-2, 2)
3. The slope is 3 and the y-intercept is (0,4)
4. The x intercept -3 and the y-intercept is 6
5. Passing through the points (-1, -2) and (5, 3)

Answers: 3

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Answer from: saysay92

A linear equation can be written in slope intercept and in standard form.

(1) Slope = 5, Point = (-1,4)

The equation is calculated as:

$\mathbf{y = m(x - x_1) + y_1}$

So, we have:

$\mathbf{y = 5(x + 1) + 4}$

$\mathbf{y = 5x + 5 + 4}$

$\mathbf{y = 5x + 9}$

Subtract 5x from both sides

$\mathbf{-5x + y = 9}$

So, the equations are:

Slope intercept: $\mathbf{y = 5x + 9}$ Standard form: $\mathbf{-5x + y = 9}$ .

(2) Points = (3,4) and (-2,2)

First, we calculate the slope (m)

$\mathbf{m =\frac{y_2 - y_1}{x_2 - x_1}}$

So, we have:

$\mathbf{m =\frac{2-4}{-2-3} = \frac{2}{5}}$

The equation is calculated as:

$\mathbf{y = m(x - x_1) + y_1}$

So, we have:

$\mathbf{y = \frac 25(x - 3) + 4}$

$\mathbf{y = \frac 25x - \frac{6}{5} + 4}$

$\mathbf{y = \frac 25x + \frac{-6 + 20}{5}}$

$\mathbf{y = \frac 25x - \frac{14}{5}}$

Multiply through by 5

$\mathbf{5y = 2x - 14}$

Rewrite as:

$\mathbf{2x - 5y = 14}$

So, the equations are:

Slope intercept: $\mathbf{y = \frac 25x - \frac{14}{5}}$ Standard form: $\mathbf{2x - 5y = 14}$ .

(3) Slope = 3, y-intercept = 4

The m and b in y = mx + b represent slope and y-intercept, respectively.

So, we have:

$\mathbf{y = 3x + 4}$

Rewrite as:

$\mathbf{3x - y = -4}$

So, the equations are:

Slope intercept: $\mathbf{y = 3x + 4}$ Standard form: $\mathbf{3x - y = -4}$ .

(4) x-intercept = -3, y-intercept = 6

Calculate the slope using:

$\mathbf{m = -\frac{y-intercept}{x-intercept}}$

$\mathbf{m = -\frac{6}{-3}}$

$\mathbf{m = 2}$

The m and b in y = mx + b represent slope and y-intercept, respectively.

So, we have:

$\mathbf{y = 2x + 6}$

Rewrite as:

$\mathbf{2x - y =-6}$

So, the equations are:

Slope intercept: $\mathbf{y = 2x + 6}$ Standard form: $\mathbf{2x - y =-6}$ .

(5) Points = (-1,-2) and (5,3)

First, we calculate the slope (m)

$\mathbf{m =\frac{y_2 - y_1}{x_2 - x_1}}$

So, we have:

$\mathbf{m =\frac{3--2}{5--1} = \frac{5}{6}}$

The equation is calculated as:

$\mathbf{y = m(x - x_1) + y_1}$

So, we have:

$\mathbf{y = \frac 56(x + 1) - 2}$

$\mathbf{y = \frac 56x + \frac 56 - 2}$

$\mathbf{y = \frac 56x + \frac{5 - 12}{6}}$

$\mathbf{y = \frac 56x - \frac{ 7}{6}}$

Multiply through by 6

$\mathbf{6y = 5x - 7}$

Rewrite as:

$\mathbf{5x - 6y =7}$

So, the equations are:

Slope intercept: $\mathbf{y = \frac 56x - \frac{ 7}{6}}$ Standard form: $\mathbf{5x - 6y =7}$

Read more about linear equations at:

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Answer from: erinxmeow8

1) The equation of the line in slope-intercept form is $y = 5\cdot x +9$ . The equation of the line in standard form is $-5\cdot x + y = 9$ .

2) The equation of the line in slope-intercept form is $y = \frac{2}{5}\cdot x +\frac{14}{5}$ . The equation of the line in standard form is $-2\cdot x +5\cdot y = 14$ .

3) The equation of the line in slope-intercept form is $y = 3\cdot x +4$ . The equation of the line in standard form is $-3\cdot x +y = 4$ .

4) The equation of the line in slope-intercept form is $y = 2\cdot x + 6$ . The equation of the line in standard form is $-2\cdot x +y = 6$ .

5) The equation of the line in slope-intercept form is $y = \frac{5}{6}\cdot x -\frac{7}{6}$ . The equation of the line in standard from is $-5\cdot x + 6\cdot y = -7$ .

Step-by-step explanation:

1) We begin with the slope-intercept form and substitute all known values and calculate the y-intercept: ( m = 5 , x = -1 , y = 4 )

$4 = (5)\cdot (-1)+b$

4 = -5 +b

b = 9

The equation of the line in slope-intercept form is $y = 5\cdot x +9$ .

Then, we obtain the standard form by algebraic handling:

$-5\cdot x + y = 9$

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

2) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = 3$ , $y_{1} = 4$ , $x_{2} = -2$ , $y_{2} = 2$ )

$3\cdot m + b = 4$ (Eq. 1)

$-2\cdot m + b = 2$ (Eq. 2)

From (Eq. 1), we find that:

$b = 4-3\cdot m$

And by substituting on (Eq. 2), we conclude that slope of the equation of the line is:

$-2\cdot m +4-3\cdot m = 2$

$-5\cdot m = -2$

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

And from (Eq. 1) we find that the y-Intercept is:

$b=4-3\cdot \left(\frac{2}{5} \right)$

$b = 4-\frac{6}{5}$

$b = \frac{14}{5}$

The equation of the line in slope-intercept form is $y = \frac{2}{5}\cdot x +\frac{14}{5}$ .

Then, we obtain the standard form by algebraic handling:

$-\frac{2}{5}\cdot x +y = \frac{14}{5}$

$-2\cdot x +5\cdot y = 14$

The equation of the line in standard form is $-2\cdot x +5\cdot y = 14$ .

3) By using the slope-intercept form, we obtain the equation of the line by direct substitution: ( m = 3 , b = 4 )

$y = 3\cdot x +4$

The equation of the line in slope-intercept form is $y = 3\cdot x +4$ .

Then, we obtain the standard form by algebraic handling:

$-3\cdot x +y = 4$

The equation of the line in standard form is $-3\cdot x +y = 4$ .

4) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = -3$ , $y_{1} = 0$ , $x_{2} = 0$ , $y_{2} = 6$ )

$-3\cdot m + b = 0$ (Eq. 3)

b = 6 (Eq. 4)

By applying (Eq. 4) on (Eq. 3), we find that the slope of the equation of the line is:

$-3\cdot m+6 = 0$

$3\cdot m = 6$

m = 2

The equation of the line in slope-intercept form is $y = 2\cdot x + 6$ .

Then, we obtain the standard form by algebraic handling:

$-2\cdot x +y = 6$

The equation of the line in standard form is $-2\cdot x +y = 6$ .

5) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = -1$ , $y_{1} = -2$ , $x_{2} = 5$ , $y_{2} = 3$ )

-m+b = -2 (Eq. 5)

$5\cdot m +b = 3$ (Eq. 6)

From (Eq. 5), we find that:

b = -2+m

And by substituting on (Eq. 6), we conclude that slope of the equation of the line is:

$5\cdot m -2+m = 3$

$6\cdot m = 5$

$m = \frac{5}{6}$

And from (Eq. 5) we find that the y-Intercept is:

$b = -2+\frac{5}{6}$

$b = -\frac{7}{6}$

The equation of the line in slope-intercept form is $y = \frac{5}{6}\cdot x -\frac{7}{6}$ .

Then, we obtain the standard form by algebraic handling:

$-\frac{5}{6}\cdot x +y =-\frac{7}{6}$

$-5\cdot x + 6\cdot y = -7$

The equation of the line in standard from is $-5\cdot x + 6\cdot y = -7$ .

Answer from: Quest

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

$\begin{bmatrix}1 & 0 & 0& |5 \\ 0& 2& 0& |8\\ 0& 0& 3& |9\end{bmatrix}$

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,

$\begin{bmatrix}1 & 0 & 0& |5 \\ 0& 1& 0& |4\\0& 0& 1& |3\end{bmatrix}$

What is the next step in solving this augmented matrix using gauss-jordan elimination method

Answer from: Quest

75%

step-by-step explanation:

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Learning Task 3. Find the equation of the line. Do it in your notebook. (a) In slope intercept form...

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