The standard form for the equation passes through known points is:

y-y1= m(x-x)

Where the slope (m)= (y2-y1)/(x2-x1)

aand (x1,y1) and (x2,y2) are points on the line:

We have the points (-1,1) and (2,3)

Then the slope m= 3-1/2-(-1)= 2/3

Then the equation is:

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

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

y= (2/3)x + 5/3

OR :

Multiply by 3:

3y -2x -5=0

The equation of the line which passes through the given points A and B is :

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

Now, we'll substitute the coordinates by their values:

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

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

We'll cross multiplying:

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

We'll open the brackets:

3y - 3 = 2x + 2

We'll move all terms to one side:

-2x + 3y - 5 = 0

The equation of the line AB is : -2x + 3y - 5 = 0 (written asĀ general form ax+by+c=0).

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

y-y1 = {(y2-y1)/(x2-x1)}(x-x1).

Therefore the line passing through A(-1,1) B(2,3) is:

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

=(2/3){x+1}

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

2x-3y+2+3 =0

2x-3y +5 = 0.