Question

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

Answer 1

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