Question

Find an equation of the line that passes through the points. (Let x be the independent variable and y be the dependent variable.)
(−3, −6) and (5, −8)

Answer 1

Answer:

$y = \dfrac{-1}{4}x + \dfrac{-27}{4}$

Step-by-step explanation:

The equation of a line is:

y = mx + b

Where:

m = slope

b = y-intercept

First thing we need to do is solve for the slope. The slope formula is:

$m = \dfrac{y_2-y_1}{x_2-x_1}$

Where:

x₁ = x-coordinate of the first point

x₂ = x-coordinate of the second point

y₁ = y-coordinate of the first point

y₂ = y-coordinate of the second point

We are given the following points:

Point 1: (-3, -6)

Point 2: (5, -8)

So let's plug in our coordinates into the slope formula:

$m = \dfrac{y_2-y_1}{x_2-x_1}\\\\ =\dfrac{(-8)-(-6)}{5-(-3)}\\\\ =\dfrac{-2}{8}\\\\ =\dfrac{-1}{4}$

So we have our new equation of this line:

$y = \dfrac{-1}{4}x + b$

What do we do then about the y-intercept?

Our points will help us out by plugging them in our equation, so we can solve for our y-intercept (b).

Let's do both to show that it would be the same:

Point 1 (-3, -6)

$y = \dfrac{-1}{4}x + b\\\\-6 = \dfrac{-1}{4}(-3) + b\\\\-6 = \dfrac{3}{4} + b\\\\-6 = \dfrac{3}{4} + b\\\\subtract \dfrac{3}{4}\;from\;both\;sides\;of\;the\;equation\\\\-6- \dfrac{3}{4}= \dfrac{3}{4}- \dfrac{3}{4}+b\\\\\\\dfrac{-24-3}{4}=0+b\\\\\\-\dfrac{27}{4} = b$

Point 2: (5, -8)

$y = \dfrac{-1}{4}x + b\\\\-8 = \dfrac{-1}{4}(5) + b\\\\-8 = \dfrac{-5}{4} + b\\\\-8 = \dfrac{-5}{4} + b\\\\subtract \dfrac{-5}{4}\;from\;both\;sides\;of\;the\;equation\\\\-8- \dfrac{-5}{4}= \dfrac{-5}{4}- \dfrac{-5}{4}+b\\\\\\\dfrac{-32-(-5)}{4}=0+b\\\\\\\dfrac{-27}{4} = b$

Now that we have b, we can insert that into the equation of the line:

$y = \dfrac{-1}{4}x + b\\\\y = \dfrac{-1}{4}x + \dfrac{-27}{4}$