Question

find the equation of line passing through (-1 1) and (2 -4)​

Answer 1

Step-by-step explanation:

equation of line passing through (-1,1) =

x + y = 0

x - y = -2

equation of line passing through (2,-4) =

2x + y = 0

2x + y = 0x -2y = 10

Answer 2

Question: $\:\:\:\:\:\:$Find the equation of line passing through (-1,1) and (2,-4).

To Find: $\:\:\:\:\:\:$Equation of the line = ?

Solution:

Let us consider that,

$\large\sf{x_1 = -1 ,x_2 = 2}$

$\large \sf{y_1 = 1 , y_2 = -4}$

We know that,

$\LARGE \: \: \: \: \: \: \: \: \bullet \: \: \sf{\frac{y - y_1}{x - x_1} =\frac{y_2- y_1}{x_{2} - x_1}}$

Applying the values,

$\: \: \: \: \: \: \: \: \: \: \: \longrightarrow \large \sf{ \frac{y - 1}{x - ( -1 )} = \frac{( - 4) - 1}{2 - ( - 1)} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \longrightarrow \large \sf{ \frac{y - 1}{x + 1} = \frac{ - 4 - 1}{2 + 1} }$

$\: \: \: \: \: \: \: \: \: \: \: \longrightarrow \large \sf{ \frac{y - 1}{x + 1} = \frac{- 5}{3}}$

While doing the cross multiplication, we get:

$\: \: \: \: \: \: \: \: \: \: \: \: \longrightarrow\sf{3(y-1)=-5(x+1)}</p><p> \\ \: \: \: \: \: \: \: \: \: \: \: \: \longrightarrow\sf{3y-3=-5x-5} \\\: \: \: \: \: \: \: \: \: \: \: \: \longrightarrow\sf{3y - 5x - 3 + 5= 0} \\ \: \: \: \: \: \: \: \: \: \: \: \: \longrightarrow\sf{ - 5x + 3y+ 2 = 0} \\ \: \: \: \: \: \: \: \: \: \: \: \: \longrightarrow\sf{5x - 3y - 2 = 0} \\ \\ \bf{ \therefore \: the \: required \: answer \: is \: 5x - 3y - 2 = 0}$