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3 years ago

Write the point slope form of an equation of the line that passes through (2,1) and (4,-3)

Mathematics

2 answers:

steposvetlana [31]3 years ago

5 0

Answer: y+3=-2(x-4)

Step-by-step explanation:

First, let’s find the slope. To do that just do (y2-y1)/(x2-x1). Substitute: (-3-2)/(3-2)= -5.

second, we want to put it in y=mx+b form so choose which coordinates you want to work with. Let’s say (2,1). So, 1=-5(2)+b

*(the one represents the y value, the -5 is slope and 2 is the x value)*

now just solve for b. b= 11

so the equation is: y=-5x+11

now, you can solve for the choices (I mean put them in y=Mx+b form) or just substitute in the coordinates to see which equation will work. In this case for choice w if we do 1+3=-2(2-4), that will be the answer.

dolphi86 [110]3 years ago

3 0

Answer:

Step-by-step explanation:

1. find the slope

y2-y1/x2-x1

-3-1=-4

4-2=2

=-2-------this is our slope

point slope equation

y-y1=m(x-x1)

y+3=-2(x-4)

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Assume the equality holds for n = N, so that

$\displaystyle \sum_{k=1}^N \frac1{k(k+1)} =1 - \frac1{N+1}$

We want to use this to establish equality for n = N + 1, so that

$\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac1{N+2}$

We have

$\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = \sum_{k=1}^N \frac1{k(k+1)} + \frac1{(N+1)(N+2)}$

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$\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac{N+1}{(N+1)(N+2)}$

$\displaystyle \sum_{k=1}^{N+1} \frac1{k(k+1)} = 1 - \frac1{N+2}$

and this proves the claim.

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