# Find the equation of a line that is parallel to line y=5x and contains the point (1,-1).

## Question:

Find the equation of a line that is parallel to line {eq}y=5x {/eq} and contains the point {eq}(1,-1). {/eq}

## Equation of a Line:

Given line is parallel to the equation of a line. So, the slope of the given line is equal to the slope of the equation of a line.

We have to find the equation of the line at the point, by using the slope {eq}\ \displaystyle y-y_{1} = m(x-x_{1}) {/eq}.

Let us consider the given line {eq}\displaystyle y=5x {/eq} as {eq}\displaystyle y=mx + c {/eq} and the point {eq}\displaystyle (1,-1) {/eq}.

Finding the equation of a line that is parallel to the given line:

{eq}\begin{align*} \\ \displaystyle m &= 5 \\ \displaystyle (x_{1}, y_{1}) &=(1, -1) \\ \displaystyle y-y_{1} &= m(x-x_{1}) \\ \displaystyle y-(-1) &= 5(x-1) \\ \displaystyle y+1 &= 5x-5 \\ \displaystyle y &= 5x-5-1 \\ \displaystyle y &= 5x-6 \end{align*} {/eq}

Therefore, the equation of a line at the point is {eq}\ \displaystyle \mathbf{\color{blue}{ y = 5x-6 }} {/eq}. 