Question

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

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 by using this concept we will find the equation of a line passing through the given points.

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

Note: Alternatively we can find the other points by using the given equation of the line. Then by using the formula of line $\left( y-{{y}_{2}} \right)=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{2}} \right)$ passes through the points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$, we will find the equation of the line.