Show that the equation of the line passing through the origin and making an angle θ with the line y = mx + c is y/x = m ± tanθ/1 ∓ tanθ

Solution:

Let the equation of the line passing through the origin be y = m\(_1\)x. From this, m\(_1\) = y/x.

If this line makes an angle of θ with line y = mx + c , then angle θ is given by

tanθ = |(m\(_1\) - m)/(1 + m\(_1\)m)|

tanθ = |(y/x - m)/(1 + (y/x)m)|

tanθ = ± (y/x - m)/(1 + (y/x)m)

tanθ = (y/x - m)/(1 + (y/x)m) or tanθ = - [(y/x - m)/(1 + (y/x)m)]

Case I:

tanθ = (y/x - m)/(1 + (y/x)m)

tanθ + (y/x) m tanθ = y/x - m

m + tanθ = y/x (1 - m tanθ)

y/x = (m + tanθ)/(1 - m tanθ)

Case II:

tanθ = - [(y/x - m)/1 + (y/x)m]

tanθ + (y/x) m tanθ = - y/x + m

m - tanθ = y/x (1 + m tanθ)

y/x = (m - tanθ)/(1 + m tanθ)

Thus, the required line is given by y/x = m ± tanθ/1 ∓ tanθ

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NCERT Solutions Class 11 Maths Chapter 10 Exercise ME Question 13

Show that the equation of the line passing through the origin and making an angle θ with the line y = mx + c is y/x = m ± tanθ/1 ∓ tanθ.

Summary:

We proved that the equation of the line passing through the origin and making an angle θ with the line y = mx + c is y/x = m ± tanθ/1 ∓ tanθ