The equation of the line is :

y-y1= m(x-x1)

(x1,y1) is any point passes through the line:

We will use A(2,4):

==> y-4=m(x-2)

==> y= m(x-2) + 4

The slope (m) = (y2-y1)/(x2-x1)

= 3-4/ -1-2= -1/-3=1/3

==> y= (1/3)(x-2) +4

==> y= (1/3)x -2/3 + 4

==> y= (1/3)x +10/3

A line which passes through 2 given points, A(2,4) and B(-1,3), has the following equation:

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

(-1-2)/(x-2) = (3-4)/(y-4)

-3/(x-2) = -1/(y-4)

We'll cross multiply:

-3(y-4) = -1(x-2)

-3y+12 = -x + 2

We'll isolate -3y to the left side. For this reason, we'll subtract 12 both sides:

-3y = -x + 2 - 12

-3y = -x - 10

We'll divide by -3:

**y = x/3 + 10/3**

We assume y = mx+c is the solution and passes through A(2,4) and B(-1,3).

Since the the point A and B lie on the line y = max+c or mx+c = y, the coordinates of the the points should satisfy the equation.

So for point (2,4), put x=2 and y = 4 in mx+c = y :

m(2)+c = 4

2m+c = 4.........(1)

Similarly for (-1,3), we get: m(-1)+c = 3

-m+c = 3.........(2).

We solve for m and from equations (1) and (3):

Eq(1) - Eq(2) eliminates c:

(2m+c)-(-m+c) = 4- 3

3m = 1

m = 1/3.

Substitute m= 1/3 in eq(1) and silve for c: 2(1/3)+c = 4. Or

c = 4-2/3 = 10/3.

So y = x/3+10/3, or 3y =x+10 , or x-3y+10 = 0 is the line that passes through A(2,4) and B(-1,3)