Solveeit Logo
Login

Question

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

How do you write an equation of a line given (5,−2)\left( 5,-2 \right) parallel to line 3y+7x=−83y+7x=-8?

Explanation

Solution

As we know that the general equation of a line is given by y=mx+cy=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 (5,−2)\left( 5,-2 \right) and parallel to 3y+7x=−83y+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+cy=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=−83y+7x=-8.
So first convert it into general form by dividing the whole equation by 3 then we will get
⇒3y3+7x3=−83\Rightarrow \dfrac{3y}{3}+\dfrac{7x}{3}=\dfrac{-8}{3}
Now, simplifying the above obtained equation we will get
⇒y=−7x3−83\Rightarrow y=-\dfrac{7x}{3}-\dfrac{8}{3}
Now, comparing the equation with the general equation we will get
⇒m=−73,y=−83\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=−73m=\dfrac{-7}{3}.
Now, the general equation of the line will be
⇒y=−73x+c\Rightarrow y=\dfrac{-7}{3}x+c
Now, the line is going through the point (5,−2)\left( 5,-2 \right).
So let us substitute x=5x=5 and y=−2y=-2 in the above equation then we will get
⇒−2=−73×5+c\Rightarrow -2=\dfrac{-7}{3}\times 5+c
Now, simplifying the above obtained equation we will get
⇒−2=−353+c ⇒−2+353=c ⇒−6+353=c ⇒293=c ⇒c=293 \begin{aligned} & \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{aligned}
So the equation of the line with slope −73\dfrac{-7}{3} and y-intercept 293\dfrac{29}{3} will be
⇒y=−73x+293 ⇒3y=−7x+29 \begin{aligned} & \Rightarrow y=\dfrac{-7}{3}x+\dfrac{29}{3} \\\ & \Rightarrow 3y=-7x+29 \\\ \end{aligned}
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 −1-1.