Question

Which is an equation of the line that passes through (–1, –5) and (–3, –7)?

Answer 1

For this case we have that by definition, an equation of the line of the slope-intersection form is given by:

$y = mx + b$

Where:

m: It's the slope

b: It is the cut-off point with the y axis

We have two points through which the line passes, then we find the slope:

$(x1, y1): (- 1, -5)\\(x2, y2): (- 3, -7)$

$m = \frac {y2-y1} {x2-x1} = \frac {-7 - (- 5)} {- 3 - (- 1)} = \frac {-7 + 5} {- 3 + 1} = \frac {-2} {- 2} = 1$

Then, the equation is of the form:

$y = x + b$

We substitute a point and find b:$-5 = -1 + b\\-5 + 1 = b\\b = -4$

Finally we have:

y = x-4

Answer:

Option B

Answer 2

Answer: Option b)

$y=x-4$

Step-by-step explanation:

The equation of a line in the intersecting slope form has the following form:

$y=mx+b$

Where

m is the slope

$m=\frac{y_2-y_1}{x_2-x_1}$  Where $(x_1, y_1)$ and $(x_2, y_2)$  are two points that belong to the line

$b=y_1-mx_1$

In this case the points are:  (–1, –5) and (–3, –7)

So:

$m=\frac{-7-(-5)}{-3-(-1)}$

$m=\frac{-7+5}{-3+1}$

$m=\frac{-2}{-2}$

$m=1$

Therefore

$b=-5-1*(-1)$

$b=-5+1$

$b=-4$

Finally the equation of the line is:

$y=x-4$