Question

On a coordinate plane, a line goes through (negative 5, negative 4) and (0, negative 3). A point is at (negative 2, 2). What is the equation of the line that is parallel to the given line and passes through the point (–2, 2)? y = One-fifthx + 4 y = One-fifthx + Twelve-fifths y = –5x + 4 y = –5x + Twelve-fifths

Answer 1

Answer:

(B) y=1/5x +12/5

Step-by-step explanation:

Answer 2

Answer:

Second option: $y=\frac{1}{5}x+\frac{12}{5}$

Step-by-step explanation:

The equation of the line in Slope-Intercept form is:

$y=mx+b$

Where "m" is the slope and "b" is the y-intercept.

We know that a line goes through the points $(-5,-4)$ and $(0,-3)$

Since the line intersects the y-axis when $x=0$, then the y-intercept of this line is:

$b=-3$

Substitute the y-intercept and coordinates of the point  $(-5,-4)$ into the equation $y=mx+b$ and solve for "m":

$-4=m(-5)-3\\\\-4+3=-5m\\\\m=\frac{-1}{-5}\\\\m=\frac{1}{5}$

By definition, the slopes of parallel lines are equal, then the slope of the other line is:

$m=\frac{1}{5}$

Knowing that it passes through the point $(-2, 2)$, we can substitute values into the equation  $y=mx+b$ and solve for "b":

 $2=\frac{1}{5}(-2)+b\\\\2+\frac{2}{5}=b\\\\b=\frac{12}{5}$

Therefoe, the equation of this line is:

 $y=\frac{1}{5}x+\frac{12}{5}$