Question

How do I write the equation of the line passing through (-3, 5) and parallel to the line y = 2x + 6.

Answer 1

parallel

y=2x+11

Step-by-step explanation

Step 1

find the equation of the line:

two lines are parellel if the slope is the same for both,so

let

`y_1=2x+6`

this function is written in slope-intercept form y=mx+b, where m is the slope

`\begin{gathered} y=2x+6\rightarrow y=mx+b \\ \text{hence} \\ slope_1=m_1=2 \\ so \\ slope_1=slope_2=m_2=2 \end{gathered}`

it means the slope of the line we are looking for is 2

Step 2

now,let's find the equation of the line

to do that, we can use this formula:

`\begin{gathered} y-y_1=m(x-x_1) \\ \text{where m is the slope} \\ \text{and (x}_1,y_1)\text{ is a point from the line} \end{gathered}`

then,let

`\begin{gathered} \text{slope}=2 \\ \text{passing point=(-3,5)} \end{gathered}`

Now,replace in the formula and solve for y

`\begin{gathered} y-y_1=m(x-x_1) \\ y-5=2(x-(-3)) \\ y-5=2(x+3) \\ y-5=2x+6 \\ \text{add 5 in both sides} \\ y-5+5=2x+6+5 \\ y=2x+11 \end{gathered}`

so, the answer is

`y=2x+11`

I hope this helps you