Question

find the equation of the line parallel to y = 2x + 2 that includes the point (3, -1). give your answer in point - slope form. y - ? = (x - ) point - slope form: y - y₁ = m(x - x₁)

Explanation:

Step1: Determine the slope (m)

Parallel lines have the same slope. The given line is \( y = 2x + 2 \), which is in slope - intercept form \( y=mx + b \) where \( m \) is the slope. So the slope of the given line is \( m = 2 \). Therefore, the slope of the line we want to find (which is parallel to the given line) is also \( m=2 \).

Step2: Identify the point \((x_1,y_1)\)

We are given that the line passes through the point \( (3,-1) \). In the point - slope form \( y - y_1=m(x - x_1) \), \( x_1 = 3 \) and \( y_1=-1 \).

Step3: Substitute into point - slope form

Substitute \( m = 2 \), \( x_1 = 3 \) and \( y_1=-1 \) into the point - slope formula \( y - y_1=m(x - x_1) \). We get \( y-(-1)=2(x - 3) \), which simplifies to \( y + 1=2(x - 3) \). But in the given format \( y-\square=\square(x - \square) \), since \( y_1=-1 \), \( y-(-1)=y + 1=y-(-1) \), so the first box (for \( y_1 \)) is \( - 1 \), the slope \( m = 2 \), and the third box (for \( x_1 \)) is \( 3 \).

Answer:

\( y - (-1)=2(x - 3) \) (or in the boxed form as per the question's format: \( y - \boxed{-1}= \boxed{2}(x - \boxed{3}) \))