The given equation of a line is
y = mx + c ... (1)
Slope of (1) is m
Let n be the slope of the required line, then tan θ = |

n−m

1+nm

|
⇒ tan θ = ± (

n−m

1+nm

)
Case (i) : When tan θ =

n−m

1+nm

Then tanθ + nm tanθ = n – m ⇒ nm tanθ – n = – m – tanθ
⇒ n =

m+tanθ

1−mtanθ

∴ Equation of the required line through the origin is y – 0 = n (x – 0)
i.e., y = (

m+tanθ

1−mtanθ

)x ⇒

y

x

=

m+tanθ

1−mtanθ

... (2)
Case (ii) : When tan θ = − (

n−m

1+mn

)
Then tanθ + mn tanθ = – n + m ⇒ n(1 + m tanθ) = m – tanθ
⇒ n =

m−tanθ

1+mtanθ

Hence, from (2) and (3), equation of line is

y

x

=

m±tanθ

1∓mtanθ

.
Hence proved.