Question

find the equation of the pine perpendicular to thr line x+y+2=0 and passing thus the point (-1, 0) ​

Answer 1

Answer:-

we have to find:

Equation of the line perpendicular to x + y + 2 and passing through the point ( - 1 , 0).

We know that,

Equation of the line perpendicular to ax + by + c = 0 and passing through the point (x₁ , y₁) is b(x - x₁) - a(y - y₁) = 0

So,

Let,

• a = 1

• b = 1

• c = 2

• x₁ = - 1

• y₁ = 0

Hence,

Required equation of the line is :

⟹ 1 [ x - ( - 1) ] - 1( y - 0 ) = 0

⟹ x + 1 - y = 0

⟹ x - y + 1 = 0

∴ Required equation of the line is x - y + 1 = 0.

___________________________

Some Important formulae :-

• Equation of a line parallel to ax + by + c = 0 and passing through (x₁ , y₁) is a(x - x₁) + b(y - y₁) = 0.

• By slope point form of a line , the equation of the line is (x - x₁)(y₂ - y₁) = (y - y₁)(x₂ - x₁).

• The equation of a line parallel to x axis at a distance of k units from it is y = k .

• The equation of a line parallel to y axis at a distance of k units from it is x = k.

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

☆ Given equation of line, l is x + y + 2 = 0------'(1)

☆ So, slope of line, l is

$\tt \:  \longrightarrow \: = - \dfrac{ - coefficient \: of \: x}{coefficient \: of \: y} \\ \tt \:  \longrightarrow \: = \: - \: \dfrac{1}{1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\tt \:  \longrightarrow \: = \: - \: 1$

☆ Let L be the required line perpendicular to (1).

☆ Let 'm' be the slope of line L.

☆ Since L is perpendicular to line (1).

$\tt\implies \:Slope \: of \: L \: (m) \: = - \dfrac{1}{Slope \: of \: l}$

$\tt\implies \:Slope \: of \: L \: (m) = \dfrac{ - 1}{ - 1} = 1$

☆ Hence, the equation of line L, which passes through (-1, 0) and having slope (m) = 1 is given by

$\tt \:  \longrightarrow \: y -0 = 1(x - ( - 1))$

$\tt \:  \longrightarrow \: y = x + 1$