Input the equation of the given line in standard form. The line including (3, 1) and (-2, 3).?

How to get the equation of the line that passes through A (3 ; 1) B (- 2 ; 3) ?

The typical equation of a line is: y = mx + b → where m: slope and where b: y-intercept


To calculate m

m = (yB - yA) / (xB - xA) = (3 - 1) / (- 2 - 3) = - 2/5

The equation of the line (AB) becomes: y = - (2/5).x + b


To calculate b

The line passes through A, so the coordinates of this point must verify the equation of the line.

y = - (2/5).x + b

b = y + (2/5).x → you substitute x and y by the coordinates of the point A (3 ; 1)

b = 1 + [(2/5) * 3]

b = 1 + (6/5)

b = 11/5


→ The equation of the line (AB) is: y = - (2/5).x + (11/5)

y = (- 2x + 11)/5

5y = - 2x + 11

2x + 5y = 11

Apply two-point form of equation of straight line :
(y - 1)/(x - 3) = (3 - 1)/(-2 - 3)
(y - 1)/(x - 3) = -2/5
5(y - 1) = -2(x - 3)
5y - 5 = -2x + 6

The required equation in standard form :
2x + 5y - 11 = 0

Use the coordinates of the points to find the slope of the line.
rise = 3 - 1 = 2
run = -2 - 3 = -5
slope = rise/run = -⅖

point-slope equation of line:
y-1 = -⅖(x-1)
You can convert it to standard form.