Question

Find the equation of a line passing through the point (–1, 2) and parpendicular to x – y + 2 = 0.​

Answer 1

Solution :-

Line is ⟂ to x - y + 2 = 0

→ x - y + 2 = 0

→ y = x + 2

Comparing it with General equation of line y = mx + c ,

we get,

→ m (Slope of line) = 1

Now, we know That, when Lines are ⟂

Than,

→ m₁ * m₂ = (-1)

So,

→ m₂ = (-1) .

_________

Hence,

→ Equation of line = mx + c

→ y = (-1)x + c

→ y = - x + c

→ y + x = c

__________

Now, it is also given that, This line passes Through (-1, 2)

So ,

→ 2 + (-1) = c

→ 2 - 1 = c

→ c = 1 .

_________

∴ Required Equation of Line is x + y - 1 = 0 .

_____________________________

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Answer 2

Question :-

Find the equation of a line passing through the point (–1, 2) and parpendicular to x – y + 2 = 0. ?

Concept :-

1) when two lines are perpendicular product of their slope is Equal to 1.

2) General equation of line is y = mx + c.

Solution :-

=> x - y + 2 = 0

=> x - y + 2 = 0

=> y = x + 2

=> so, slope of line = 1

by concept (1)

=> m₁ × m₂ = (-1)

=> 1 × m₂ = (-1)

=> m₂ = (-1)

So, our equation of line is y = mx + c

=> y = (-1)x + c

=> y = - x + c

=> y + x = c

Now, we have given that, This line passes Through (-1, 2)

put value

=> 2 + (-1) = c

=> 2 - 1 = c

=> c = 1

=> Equation of Line = x + y - 1 = 0 .