Crucial Points:

Example 1: Two points. A(3,2) and B(6,4). Find the
expression for the line that connects A to B. Solution: We will find the gradient first then use expression (1). m = [42] / [63] m = 6 / 3 m = 2 . Pick any point and apply expression (1). Here, we randomly pick point A. y  2 = (2)( x  3 ) y = 2x + 6 + 2 y = 2x + 8. 
Example 2: Two points. Line L1 passes through the origin and
(2,3). Find the linear expression for L1. Solution: The point origin is simply (0,0). This is thus a problem with two points and we will follow similar method as in previous Example 1. m = [30] / [20] m =  3/2 y  0 = (3/2) [ x 0 ] ; the origin (0,0) is an excellent choice because it simplifies the calculation although (2,3) can also be used. y =  (3/2)x. 
Example 3: A slope and a point. Line L1 has gradient
2 and passes through (5,3). Find the linear expression for L1. Solution: We can simply apply expression (2). y  3 = 2 (x  (5) ) y = 2x + 10 + 3. y = 2x + 13. 
Example 4: A rule and a point. Line L1 passes through (4,
2) and is parallel to a straight line that has gradient 3. Find the
linear expression for L1. Solution: Since L1 is parallel to a line that has gradient 3 then m1 = 3. We now apply expression (2). y  (2) = 3 (x  4) y = 3x + 12  2 y = 3x + 10. 
Example 5: A rule and a point. Line L1 passes through (1,2)
and is perpendicular to L2 that has gradient 1/4. Find the linear expression
for L1. Solution: Since L1 is perpendicular to L2 then m1.m2 = 1. Thus, m1 = 1 / (1/4) and m1 = 4. y  2 = 4 ( x  1) y = 4x + 4 + 2 y = 4x + 6. 
Example 6: Problem with Sketch Find
the linear expression for L1. Solution: Note that L1 passes through the origin (0,0) and L1 is perpendicular to L2. Thus, m1 = 1/m2. We need to find m2 first. The two points on L2 are (0, 5) and (3, 0). Thus m2 = [50] / [03] m2 = 5/3. So, m1 = 1/ (5/3) m1 = 3/5. y = 3/5 x 