Question

Show that the equation of the line passing through the origin and making an angle θ with the line `y = mx + c " is " y/c = (m+- tan theta)/(1 +- m tan theta)`.

Answer 1

The equation of line PA is y = mx + c

This line makes an angle θ with OP.

Slope of line PA = m

Let slope of OP = m₁.

Now tan θ = ± `("m"_1 - "m")/(1 + "m"_1"m")`, where m = tan θ

Taking +ve sign, tan θ = ± `("m"_1 - "m")/(1 + "m"_1"m")`

or `(1 + "m"_1"m")tan θ = "m"_1 - "m"`

or tan θ + m₁ m tan θ = m₁ - m

or m + tan θ = m(1 - m tan θ)

or `"m"_1 = ("m" + tan θ)/(1 - "m" tan θ)`

Taking -ve sign,

`"m"_1 = ("m" + tan θ)/(1 - "m" tan θ)`

`(- 1 + "m"_1"m") tan θ = "m"_1 - "m"`

or (1 + m₁m) tan θ  = −m₁ + m

m₁(1 + m tan θ) = m − tan θ

∴ `"m"_1 = ("m" - tan θ)/(1 + "m" tan θ)`

Therefore, both slopes are represented by `("m" ± tan θ)/(1 "m" ± tan θ)`.

∴ The equation of the line passing through the origin (0, 0),

(y - 0) = m₁ (x - 0)

y =m₁ × x

or `"y"/"x" = "m"_1`

∴ equation of required lines

`"y"/"x" = ("m" ± tan θ)/(1 ± tan θ)`