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I am Lyosha [343]
3 years ago
14

What’s the slope of the line that passes through (3,5) and (2,6)

Mathematics
2 answers:
Aleksandr [31]3 years ago
8 0

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{6}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{6}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{2}-\underset{x_1}{3}}}\implies \cfrac{1}{-1}\implies -1

Licemer1 [7]3 years ago
7 0

Answer:

8

Step-by-step explanation:

So what we need now are the two points you gave that the line passes through. Let's call the first point you gave, (3,5), point #1, so the x and y numbers given will be called x1 and y1. Or, x1=3 and y1=5.

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

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

m=

6 - 5

2 - 3

or...

m=

1

-1

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:

(3,5). When x of the line is 3, y of the line must be 5.

(2,6). When x of the line is 2, y of the line must be 6.

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

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:

(3,5). y=mx+b or 5=-1 × 3+b, or solving for b: b=5-(-1)(3). b=8.

(2,6). y=mx+b or 6=-1 × 2+b, or solving for b: b=6-(-1)(2). b=8.

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

(3,5) and (2,6)

is

y=-1x+8

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