# Line's equation: What is the equation of the line that passes through the point (-1;1) and (0;3)?

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You need to use point slope form of equation of the line, such that:

`y - y_0 = m(x - x_0)`

`(x_0,y_0)` represents a point on line

m represents the slope of the line

Selecting `(x_0,y_0) = (0,3)` , you need to determine the slope using slope equation, such that:

`m = (y_1 - y_0)/(x_1 - x_0)`

Since the problem provides `(x_1,y_1) = (-1,1)` yields:

`m = (1 - 3)/(-1 - 0) => m = (-2)/(-1) => m = 2`

Hence, substituting 2 for m and the coordinates `x_0, y_0` , yields:

`y - 3 = 2(x - 0) => y - 3 = 2x => y = 2x + 3`

**Hence, evaluating the equation of the line, under the given conditions, yields **`y = 2x + 3.`

The equation of the line that passes through the points is:

(0+1)/(x+1) = (3-1)/(y-1)

We'll compute and we'll get:

1/(x+1) = 2/(y-1)

We'll cross multiply and we'll get:

2(x+1) = y-1

We'll remove the brackets:

2x + 2 = y - 1

We'll put the equation into the general form:

ax + by +c = 0

We'll move all terms to the left side:

2x - y + 2 + 1 = 0

We'll combine like terms:

2x - y + 3 = 0

The equation of the line is:

2x - y + 3 = 0