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2 years ago

Find the equation of a line passing through (-2/3, 5) that is parallel to 4x-2y=6. Express your answer in point-slope form.

Mathematics

1 answer:

faust18 [17]2 years ago

6 0

Using the information given, it is found that the linear function is given by:

$y - 5 = 2\left(x + \frac{2}{3}\right)$

A linear equation, in point-slope formula, is modeled by:

$y - y_1 = m(x - x_1)$

In which:

m is the slope.

The point is $(x_1, y_1)$ .

If two lines are parallel, they have the same slope.

Parallel to 4x - 2y = 6, which, in standard form is:

$2y = 4x - 6$

$y = \frac{4}{2}x - \frac{6}{2}$

$y = 2x - 3$

Hence, the slope is $m = 3$ .

Point $\left(-\frac{2}{3}, 5\right)$ , hence, the equation of the line is:

$y - y_1 = m(x - x_1)$

$y - 5 = 2\left(x + \frac{2}{3}\right)$

To learn more about linear functions in point-slope form, you can take a look at brainly.com/question/13967935

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Learning Task 3. Find the equation of the line. Do it in your notebook.

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Answer:

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

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3 years ago