Question

1. write the equation of the line passing through (3, - 1) and is parallel to the line x - 3y = 9.

x - 3y = 9, (3, - 1)

1. write the equation of the line passing through (4, 3) and is perpendicular to the line y = 2x - 4.

y = 2x - 4, (4, 3)

1. write the equation of the line passing through (- 5, 1) and is perpendicular to the line 5x + 3y = - 21.

5x + 3y = - 21, (- 5, 1)

Explanation:

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Step1: Find the slope of the given line

Rewrite $x - 3y=9$ in slope - intercept form $y = mx + b$ (where $m$ is the slope).
$-3y=-x + 9$, so $y=\frac{1}{3}x - 3$. The slope $m_1=\frac{1}{3}$. Since parallel lines have the same slope, the slope of the new line $m=\frac{1}{3}$.

Step2: Use the point - slope form

The point - slope form is $y - y_1=m(x - x_1)$, with $(x_1,y_1)=(3,-1)$ and $m = \frac{1}{3}$.
$y-(-1)=\frac{1}{3}(x - 3)$.

Step3: Simplify the equation

$y + 1=\frac{1}{3}x-1$.
$y=\frac{1}{3}x-2$ or $x - 3y=6$.

Step1: Find the slope of the perpendicular line

The slope of the line $y = 2x-4$ is $m_1 = 2$. For perpendicular lines, the product of their slopes is $- 1$. Let the slope of the new line be $m$. Then $2m=-1$, so $m=-\frac{1}{2}$.

Step2: Use the point - slope form

Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(4,3)$ and $m=-\frac{1}{2}$.
$y - 3=-\frac{1}{2}(x - 4)$.

Step3: Simplify the equation

$y-3=-\frac{1}{2}x + 2$.
$y=-\frac{1}{2}x+5$ or $x + 2y=10$.

Step1: Find the slope of the given line

Rewrite $5x + 3y=-21$ in slope - intercept form $y=mx + b$.
$3y=-5x-21$, so $y=-\frac{5}{3}x - 7$. The slope of this line is $m_1=-\frac{5}{3}$. For perpendicular lines, if the slope of one line is $m_1$ and the other is $m$, then $m_1m=-1$. So $-\frac{5}{3}m=-1$, and $m=\frac{3}{5}$.

Step2: Use the point - slope form

Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(-5,1)$ and $m=\frac{3}{5}$.
$y - 1=\frac{3}{5}(x + 5)$.

Step3: Simplify the equation

$y-1=\frac{3}{5}x+3$.
$y=\frac{3}{5}x + 4$ or $3x-5y=-20$.

Answer:

$y=\frac{1}{3}x - 2$ (or $x - 3y=6$)

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