# Find the equation for the line passes through (1,3) and (2,-1)

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The line passes through A(1,3) and B(2,-1)

Then:

y-yA = m (x-xA) where m is the slope.

m= (yB-yA)/(xB-xA) = (-1-3)/(2-1) = -4

Then:

y-3 = -4(x-1)

y-3 = -4x + 4

**==> y= -4x +7**

Let's approach the problem in this way: If a point belongs to a line, that means that the coordinates of the point verifies the equation of the line.

Let's put the equation of the line in the standard form:

y = ax + b

The point (1,3) is on the line if and only if the coordinates verify the equtaion of the line:

3 = a*1 + b (1)

The point (2,-1) is on the line if and only if the coordinates verify the equtaion of the line:

-1 = a*2 + b (2)

From (1), we'll get a = 3-b (3)

We'll re-write (2) and we'll substitute (3) in (2):

-1 = a*2 + b

2a + b = -1

2(3-b) + b = -1

We'll remove the brackets:

6 - 2b + b + 1 = 0

We'll combine like terms:

7 - b = 0

We'll add b both sides:

** b = 7**

a = 3-b

a = 3-7

**a = -4**

The equation of the line is:

**y = -4x + 7**

The equation of the line through the points (x1,y1) and (x2, y2) is

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

Thegiven points are ((1,3) and (2-1).

So substituting (x1,y1) = (1,3) and (x2,y2) = (2,-1), we get:

y-3 = {(-1-3)/(2-1)}(x-1)

y-3 = -4(x-1) = -4x+1.

4x+y-3-1 =0

4x+y-4 = 0. Is the equation that passes through (1,3) and (2,-1).

The equation of a line passing through two points ( x1, y1) and ( x2, y2) is x - x2 = [( x2 -x1)/ (y2 - y1)] * (y - y2)

Substituting the values from the points we have (1 , 3) and (2, -1)

(x -2) = [( 2 -1)/ (-1 - 3)] * (y - (-1) )

=> x - 2 =( -1 / 4) * ( y + 1)

=> 4 * (x - 2) = - y - 1

=> 4x - 8 = - y - 1

=> 4x + y - 7 =0

**The required equation is 4x + y - 7 =0**