Given the points ( 2,3) and (5,8) .

We need to find the equation of the line that passes through the given points.

We will write the equation into the standard form.

==> y-y1 = m (x-x1) where (x1,y1) is any point passes through the line and m is the slope.

We know that the slope is given by the formula:

m = (y2-y1)/(x2-x1) = ( 8-3)/(5-2) = 5/3

==> m = 5/3

==> Now we ill substitute with m = 5/3 and the point (2,3)

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

==> y-3 = (5/3)x - 10/3

==> y= (5/3)x - 10/3 + 3

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

We will multiply by 3.

==> 3y = 5x -1

**==> 3y -5x +1 = 0**

The equation of the line that passing through 2 given points is:

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

We'll identify x1 = 2, x2 = 5, y1 = 3 , y2 = 8.

We'll substitute them into the formula:

(5 - 2)/(x - 2) = (8 - 3)/(y - 3)

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

We'll cross multiply:

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

We'll remove the brackets:

5x - 10 = 3y - 9

We'll add 9 both sides:

3y = 5x - 10 + 9

3y = 5x - 1

We'll divide by 3:

y = 5x/3 - 1/3

**The equation of the line that passing through the points (2,3) and (5,8) is: **

**y = 5x/3 - 1/3**