# Given the points Pn(n,2n+1), which is the equation of the line that passes through the points P1 and P2?

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To write the equation, we'll use the formula that describes the equation of the line which passes through 2 given points:

(y2-y1)/(y-y1) = (x2-x1)/(x-x1)

But first, let's find the coordinates for P1 and P2.

P1(1,2*1+1) and P2(2,2*2+1)

P1(1,3) and P2(2,5)

Now, we'll substitute the coordinates of P1, P2, into the equation:

(5-3)/(y-3) = (2-1)/(x-1)

2/(y-3) = 1/(x-1)

We'll cross multiply and we'll get:

2*(x-1) = (y-3)

We'll open the brackets and we'll get:

2x-2 = y-3

We'll move all terms to one side and we'll get the general form of the equation of the line that passes through P1, P2:

**2x-y+1=0**

Pn(x,2n+1). To find the equation of the line that passes through P1 and P2.

Solution:

P1 has the coordinates (1, 2*1+1) = (1 , 3)

P2 has the coordinates (2, 2*2+1) = (2, 5)

The line passing through (x1,y1) and (x2,y2) is

y-y1 = {(y2-y1)/(x2-x1)}(x-x1). So the line passing through P1(1,3) and P2(2,5) Is

y-3 = [(5-3)/(2-1)](x-1) = 2(x-1) Or

y-3 = 2(x-1)

2x-y-2+3 = 0

2x-y+1 = 0.