Question

Find the equation, in standard form, of the line passing through the points (3,-4) and (5,1).?

Answer 1

Slope intercept form line equation is y = mx + c.   m is slope and c is y-intercept.

m =( y2-y1)/(x2-x1) = (1-(-4)/(5-3) = 5 / 2

the line equation is y = 5/2x + c

find the y-intercept value by substituting (5,1) in y = 5/2x + c

1 = 5/2 * 5 + c

c = 1 - 25/2 = -23/2

Substitute c = -23/2 in line equation.

the line equation is y = 5/2x - 23/2

Standard form line equation is Ax+By = c

Multiply each side by 2.

2y = 5x - 23

Subtract 5x from each side.

5x - 2 y -23 = 0

Add 23 to each side.

5x - 2y = 23

The line equation in slope-intercept form is y = 5/2x - 23/2

The line equation in standard form is 5x - 2y = 23.

Answer 2

Given points say (x1,y1) = (3,-4)

                         (x2,y2) = (5,1)

m = y2-y1/x2-x1

   =1-(-4)/5-3

   = 5/2

Standerd form of the line ⇒ y-y1 = m(x-x1)

y-(-4) = 5/2(x-3)                      (simplification)

2(y+4) = 5(x-3)

2y+8 = 5x-15

2y = 5x-15-8

2y = 5x-23

2y-5x+23 = 0

The line equation is 5x-2y-23 = 0