Question

Find the equation of the straight line which is passing through the points (1, - 1 )and (3, 5)​

Answer 1

Answer:

by distance formula

d=√(x2-x1)+(y2-y1)

let a(1,-1)and b(3,5)

=√(3-1)+(5+1)

=√2+6

=√8

=2√2

Answer 2

Answer:

3x - y - 4 = 0

Step-by-step explanation:

Let the points be A:($x_{1}$  ,  $y_{1}$) ≡ (1 , -1)

and    B:($x_{2}$ , $y_{2}$) ≡ (3, 5)

Equation of line in two point form is given by

           

(y - $y_{1}$) = ($\frac{( y_{2} - y_{1} )}{(x_{2} - x_{1} ) }$)(x - $x_{1}$)

⇒ (y - [-1]) = $\frac{(5 - [-1])}{(3 - 1)}$(x - 1)

⇒ (y + 1) = $\frac{6}{2}$(x - 1)

⇒ (y + 1) = 3(x - 1)

⇒ y + 1 = 3x - 3

⇒ 3x - y - 4 = 0

Equation is required line is

      3x - y - 4 = 0