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

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### 3 Answers

We have to find the line passing through the points (-1,1) and (0,3).

The equation of the line through the points (x1, y1) and (x2, y2) is given by y-y1 = [(y2- y1)/ (x2 - x1)]*(x - x1)

Here we have x1 = -1 , x2 = 0, y1 = 1 and y2 = 3.

Therefore the equation of the line is:

y - 1) = [( 3 -1))/(0 -(-1))]*( x -(-1))

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

=> y - 1 = 2x + 2

=> 2x - y +3 =0

**Therefore the line passing through the points (-1,1) and (0,3) is 2x - y +3 =0.**

The equation of the line passing through (x1, y1) and (x2,y2) is given by:

y-y1 = {(y2-y1)/(x2-x1)}(x-x1)

Therefore the equation of th line passing thruogh the points (1,1) and (0,3) is given bY

y -1 = {(3-1)/(0-1)}(x-1).

y-1 = -2 (x-1).

y-1 = -2x+2.

2x+y -1-2 = 0.

2x+y-3 = 0 is the line that passes through the points (1,1) and (0,3).

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**