# Determine the equation of the line that contains the point A(1;3) si B(-2;-3)

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

We need the equation for the line passes through A and B

The formula for the line:

y-ya = m(x-x1) where m is the slope:

m= (y2-y1)/(x2-x1) = -3-3/-2-1= -6/-3= 2

Now let us substitute with m=2 and the point A(1,3)

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

==> y= 2x -2 + 3

==> **y= 2x +1**

To determine the equation of the line AB, we'll write the formula:

(xB-xA)/(x-xA) = (yB-yA)/(y-yA)

Now, we'll substitute the coordinates of the points A and B, into the formula above:

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

-3/(x-1) = -6/(y-3)

We'll divide by -3 both sides:

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

We'll cross multiply:

2x - 2 = y - 3

y = 2x + 1

**So, the equation of the line AB is: y = 2x + 1**

To detrmine the equation of the line through A(1,3) and B(-2,-3)

Solution:

The line joining (x1,y1) and (y1,y2) has the equation:

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

So the equation of AB is

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

(y-3) (-2) = (-6)(x-1)

y-3 = 2(x-1)

2x-y-2+3 = 0

2x-y-1 = 0.