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Find the slope of a line parallel to the line that passes through (-4, -1) and (-1, 3)

Sagot :

Answer:

[tex] \boxed{\rm \: Slope(m)= \cfrac{4}{ 3} }[/tex]

OR

[tex] \boxed{\rm \: Slope(m) \approx \: 1.30} \rm \: (nearest \: tenth)[/tex]

Step-by-step explanation:

Given two coordinates:

  • (-4, -1) and (-1, 3)

To Find:

  • The slope of given two coordinates

Solution:

In this case,we'll need to use the slope's formula,to get the slope.

Here's the formula:

[tex] \rm \: Slope(m)= \: \cfrac{ y_2 -y_1 }{x_2 - x_1} [/tex]

  • Slope is usually denoted as m.

According to the question:

  • [tex] \rm (y_2 , y _1) = (3,-1)[/tex]
  • [tex]\rm (x_2 , x_1) = (-1,-4)[/tex]

Substitute them accordingly:

[tex] \rm \: Slope(m)= \cfrac{3 - ( - 1)}{ - 1 - ( - 4)} [/tex]

Simplify then, using a rule called PEMDAS.

[tex] \rm \: Slope(m)= \cfrac{3 + 1}{ - 1 +4 } [/tex]

[tex] \rm \: Slope(m)= \cfrac{4}{ 3} [/tex]

[tex] \rm \: Slope(m) \approx \: 1.30[/tex]

Hence, we can conclude that:

  • The slope of a line parallel to the line that passes through (-4, -1) and (-1, 3) is 4/3 in exact form and 1.30 in approximate decimal form.
View image Itz5151

Hi student, let me help you out! :)

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We are asked to find the slope of a line parallel to the line that passes through

[tex]\star~\mathrm{(-4,-1)}\star[/tex]

[tex]\star~\mathrm{(-1,3)}[/tex]

[tex]\triangle~\fbox{\bf{KEY:}}[/tex]

  • The slope formula will help us out! :)

Here's the formula:

[tex]\star\boxed{\mathrm{\cfrac{y2-y1}{x2-x1}}}[/tex]

y2 and y1 are y-coordinates

x2 and x1 are x-coordinates

Substitute the given values:

[tex]\star\mathrm{\cfrac{3-(-1)}{-1-(-4)}}[/tex]

Simplify!

[tex]\star\mathrm{\cfrac{3+1}{-1+4}}[/tex]

Simplify more!

[tex]\star\mathrm{\cfrac{4}{3}}[/tex]

Now, what is the slope of the line parallel to this one?

Well, parallel lines have the same slope.

[tex]\star \ \mathrm{The \ slope \ of \ the \ line \ parallel \ to \ this \ one \ is \ \cfrac{4}{3}}.[/tex]

Hope it helps you out! :D

Ask in comments if any queries arise.

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~Just a smiley person helping fellow students :)