Question 30494
Find the equation, in standard form, of the line that passes through the points (5,6) and (7, 3)
Given two points P(x1,y1) and Q(x2,y2) on a given line
the equation to the given line is 
(y-y1)/(y1-y2) = (x-x1)/(x1-x2)----(1)
Therefore given two points P(5,6) and Q(7, 3), 
the equation to the line passing through P and Q is  
(here x1 = 5, y1 = 6; x2 = 7, y2 = 3 )
(y-6)/(6-3) = (x-5)/(5-7)
That is (y-6)/3 = (x-5)/(-2)
Multiplying by 6 on both the sides  (since lcm of 3 and 2 is 6)
2(y-6) = -3(x-5)
2y-12 = -3x+15
3x+2y-12-15 = 0 (grouping like terms,changing sign while changing side)
3x+2y-27 =0   ----(*)
which is the equation to the line in the general form
If by standard form you mean the slope and the y-intercept form
retain the y-term on the left side and take the other terms to the right 
and then divide by the coefficient 2 of y
then the slope and y-intercept form of the required line is 
y = (-3/2)x +(27/2) ----(**)
Verification: Testing for P(5,6) and Q(7, 3)in (*)
LHS = 3x+2y-27 
= 3X(5)+2X(6)-27  (putting x = 5 and y = 6 in (1) )
= 13+12-27
= 27-27
= 0
=RHS 
P(5,6) is a point on (*)

Now putting x = 7 and y = 3 in (*) 
that is checking if Q(7,3) is a point on (*)
LHS = 3x+2y-27 
= 3X(7)+2X(3)-27  
= 21+6-27
= 27-27
= 0
=RHS 
This implies Q(7,3) is a point on (*)