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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)`.
Solution
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)`
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