Question

Consider the line - 3x +9y = 5.
Find the equation of the line that is perpendicular to this line and passes through the point (5, -1).
Find the equation of the line that is parallel to this line and passes through the point (5, -1).

Answer 1

Answer:

$\textsf{Perpendicular to given line}:y=-3x+14$

$\textsf{Parallel to given line}: y=\dfrac13x-\dfrac83$

Step-by-step explanation:

Rewrite the given equation to make y the subject:

$\begin{aligned}-3x+9y &=5 \\ \implies 9y &=3x+5 \\ \implies y &=\dfrac13x+\dfrac59\end{aligned}$

Therefore, the slope of the given equation is $\frac13$.

If two lines are perpendicular to each other, the product of their slopes will be -1.  Therefore, the slope (m) of the line that is perpendicular to the given line is:

$\begin{aligned}m \times \dfrac13 & =-1\\ \implies m & =-3\end{aligned}$

To find the equation of the line, substitute the found slope (-3) and the point (5, -1) into the point-slope form of a linear equation:

$\begin{aligned}y-y_1 & =m(x-x_1)\\ \implies y-(-1) &=-3(x-5) \\ y+1 & =-3x+15 \\ y &=-3x+14\end{aligned}$

If two lines are parallel to each other, their slopes will be the same.  Therefore, the slope (m) of the line that is parallel to the given line is $\frac13$

To find the equation of the line, substitute the slope ($\frac13$) and the point (5, -1) into the point-slope form of a linear equation:

$\begin{aligned}y-y_1 & =m(x-x_1)\\\\ \implies y-(-1) &=\dfrac13(x-5) \\\\ y+1 & =\dfrac13x-\dfrac53 \\\\ y &=\dfrac13x-\dfrac83\end{aligned}$

Answer 2

• -3x+9y=5
• 9y=3x+5
• y=1/3x+5/9

Slope =m=1/3

Note that perpendicular lines have slopes negative reciprocal to each other .

• Slope of the perpendicular line=-3

Equation in point slope form

• y+1=-3(x-5)
• y+1=-3x+15
• y=-3x+14

And

parallel lines have equal slope

Equation of parallel line

• y+1=1/3(x-5)
• 3y+3=x-5
• 3y=x-8
• y=x/3-8/3