Question

Find the equation in standard form of the line passing through the points ( 2,-3) and (4,2)

Answer 1

The "equation of a straight line in standard form" is either
Ax+By+C=0 or Ax+By=D.

Method 1:  Find the slope of the line connecting the 2 given points, (2,-3) and (4,2).  It is m=5/2.

Write the slope-intercept form of the associated equation of a straight line, y=mx+b:  y=(5/2)x+b

Now substitute the coordinates of the other given point (4,2) into 
y=(5/2)x+b, to obtain a value for b:

2=(5/2)(4)+b      becomes 2=10+b.  Thus, b=-8.

Thus, our equation y=mx+b becomes y=(5/2)x-8.  But we must re-write this in "standard form."  To do this:  Multiply all terms of this equation by 2 to eliminate fractions:

2y=(5/2)(2)x-2(8), or 2y=5x-16.

Rearranging (into "standard form):  16=5x-2y, or 5x-2y=16.

Be certain to check this result by substituting the coordinates of one or both of the given points.  If the equation is true after these coordinates have been subst., then you'll know the equation is correct.

5(2)-2(-3)=16

10+6=16  (true, so 5x-2y=16 is correct.

Answer 2

Slope = 5/2
b = -8
Slope-int.: y = 5/2x -8

Standard: 5x - 2y = 16