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77julia77 [94]
3 years ago
8

Which is an equation of the line that passes through (2,-5) and (6, 3)?

Mathematics
2 answers:
Shkiper50 [21]3 years ago
8 0

Answer:

Step-by-step explanation:

You want to find the equation for a line that passes through the two points:

(-2,-5) and (6,3).

First of all, remember what the equation of a line is:

y = mx+b

Where:

m is the slope, and

b is the y-intercept

First, let's find what m is, the slope of the line...

The slope of a line is a measure of how fast the line "goes up" or "goes down". A large slope means the line goes up or down really fast (a very steep line). Small slopes means the line isn't very steep. A slope of zero means the line has no steepness at all; it is perfectly horizontal.

For lines like these, the slope is always defined as "the change in y over the change in x" or, in equation form:

So what we need now are the two points you gave that the line passes through. Let's call the first point you gave, (-2,-5), point #1, so the x and y numbers given will be called x1 and y1. Or, x1=-2 and y1=-5.

Also, let's call the second point you gave, (6,3), point #2, so the x and y numbers here will be called x2 and y2. Or, x2=6 and y2=3.

Now, just plug the numbers into the formula for m above, like this:

m=  

3 - -5

6 - -2

or...

m=  

8

8

or...

m=1

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

y=1x+b

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

(-2,-5). When x of the line is -2, y of the line must be -5.

(6,3). When x of the line is 6, y of the line must be 3.

Because you said the line passes through each one of these two points, right?

Now, look at our line's equation so far: y=1x+b. b is what we want, the 1 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-2,-5) and (6,3).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!.

You can use either (x,y) point you want..the answer will be the same:

(-2,-5). y=mx+b or -5=1 × -2+b, or solving for b: b=-5-(1)(-2). b=-3.

(6,3). y=mx+b or 3=1 × 6+b, or solving for b: b=3-(1)(6). b=-3.

See! In both cases we got the same value for b. And this completes our problem.

The equation of the line that passes through the points

(-2,-5) and (6,3)

is

y=1x-3

vladimir2022 [97]3 years ago
6 0

Answer:

y = 2x - 9

Step-by-step explanation:

m = (y2 -y1)/(x2-x1)

m = ( 3-(-5))/ ( 6 - 2)

m = 8/4 = 2

y = mx + b

3 = 2( 6 )+ b

b = 3- 12 = -9

y = 2x - 9

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3 years ago
Find the perimeter of a triangle with vertices A(3, 1), B(2, -1), and C(-3, 2). Leave
natulia [17]

The perimeter of triangle is: 14.15 units

Step-by-step explanation:

First of all we have to find the lengths of sides of triangles

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A(3, 1), B(2, -1), and C(-3, 2)

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d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\

So,

AB = \sqrt{(2-3)^2+(-1-1)^2}\\=\sqrt{(-1)^2+(2)^2}\\=\sqrt{1+4}\\=\sqrt{5}\\=2.24\ unitsBC = \sqrt{(-3-2)^2+(2+1)^2}\\=\sqrt{(-5)^2+(3)^2}\\=\sqrt{25+9}\\=\sqrt{34}\\= 5.83\ units\\AC = \sqrt{(-3-3)^2+(2-1)^2}\\=\sqrt{(-6)^2+(1)^2}\\=\sqrt{36+1}\\=\sqrt{37}\\=6.08\ units

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Hence,

The perimeter of triangle is: 14.15 units

Keywords: Triangle Perimeter

Learn more about perimeter at:

  • brainly.com/question/573729
  • brainly.com/question/572693

#LearnwithBrainly

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sashaice [31]

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