Input in standard form the equation of the given line. The line that passes through (-1, -3) and (2, 1)

Question

 Need the standard form of the line.

Answer 1

The equation of line is has a slope and a y-intercept.  It is usually in the form  y=mx+b.

 

 

First, we need to find the slope of this line using the coordinates.

 

m = (y₂ - y₁) / (x₂ - x₁)

m = (1 - (-3)) / (2 - (-1))

m = 4 / 3

 

 

The equation of the line so far is

 

y = (4/3)x + b

 

Next, we plug in the values of any of the coordinates into the equation to find b.  Lets use (2, 1).

 

1 = (4/3)(2) + b

 

1 = (8/3) + b

 

-5/3 = b

 

 

The equation of the line is

 

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

 

 

Next, we need to get the standard form. The equation of any line in this form is  

Ax + Bx = C. 

 

A, B, and C are constants.

 

Multiply both sides of the equation by 3.

 

3y = 4x - 5

 

 

Subtract 4x on both sides of equation.

 

-4x + 3y = -5

 

Answer 2

Standard form is ax + by = c

 

We have two points (-1,-3), (2,1) so lets use the point slope formula first  y-y₁=m(x-x₁) where

(x₁,y₁) is a point on the line.   First we need to find the slope m

 

To find slope we use m= (y₂-y₁)/(x₂-x₁) = (1-(-3))/(2-(-1)) = (1+3)/(2+1) = 4/3  so m = 4/3

 

Use the slope m=4/3 with either original point to plug into the point slope formula.  I'll use (2,1)

 

y-1 = (4/3)(x-2)

y-1 = (4/3)x - 8/3

Put into standard form

(4/3)x - y = -1 + 8/3

 

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

 

This is a correct answer for the problem; but if you wish to eliminate the fractions

you can multiply through by 3 giving:

 

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

 

4x-3y = 5