Question

How do you write an equation of a line given $\left( 5,-2 \right)$ parallel to line $3y+7x=-8$?

Answer 1

Hint: As we know that the general equation of a line is given by $y=mx+c$, where m is the slope of line and c is the y-intercept of the line. So first we will find the slope and then by using the slope intercept form we will find the y-intercept then by using both values we get the equation of the line.

Complete step by step answer:
We have been given that a line is going through $\left( 5,-2 \right)$ and parallel to $3y+7x=-8$.
We have to find the equation of the line.
Now, we know that the slope intercept form of a line is given as $y=mx+c$, where m is the slope of line and c is the y-intercept of the line.
Now, we have given the equation of the line is $3y+7x=-8$.
So first convert it into general form by dividing the whole equation by 3 then we will get
$\Rightarrow \dfrac{3y}{3}+\dfrac{7x}{3}=\dfrac{-8}{3}$
Now, simplifying the above obtained equation we will get
$\Rightarrow y=-\dfrac{7x}{3}-\dfrac{8}{3}$
Now, comparing the equation with the general equation we will get
$\Rightarrow m=\dfrac{-7}{3},y=\dfrac{-8}{3}$
Now, both the lines are parallel. It means they have the same slope so the slope of the line will be $m=\dfrac{-7}{3}$.
Now, the general equation of the line will be
$\Rightarrow y=\dfrac{-7}{3}x+c$
Now, the line is going through the point $\left( 5,-2 \right)$.
So let us substitute $x=5$ and $y=-2$ in the above equation then we will get
$\Rightarrow -2=\dfrac{-7}{3}\times 5+c$
Now, simplifying the above obtained equation we will get
$\begin{align}
  & \Rightarrow -2=\dfrac{-35}{3}+c \\
 & \Rightarrow -2+\dfrac{35}{3}=c \\
 & \Rightarrow \dfrac{-6+35}{3}=c \\
 & \Rightarrow \dfrac{29}{3}=c \\
 & \Rightarrow c=\dfrac{29}{3} \\
\end{align}$
So the equation of the line with slope $\dfrac{-7}{3}$ and y-intercept $\dfrac{29}{3}$ will be
$\begin{align}
  & \Rightarrow y=\dfrac{-7}{3}x+\dfrac{29}{3} \\
 & \Rightarrow 3y=-7x+29 \\
\end{align}$
Hence above is the required equation of line.

Note: The point to be remembered is that parallel lines in standard form have the same coefficients of x and y. Also parallel lines have the same slope whereas the product of slopes of two lines perpendicular to each other is $-1$.