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Question
Prove that the lines x = py + q, z = ry + s and x = p′y + q′, z = r′y + s′ are perpendicular if pp′ + rr′ + 1 = 0.
Solution
Given that, x = py + q
⇒ y = `(x - "q")/"p"`
And z = ry + s
⇒ y = `(z - "s")/"r"`
∴ The equation becomes `(x - "q")/"p" = y/1 = (z - "s")/"r"` in which d’ratios are a1 = p, b1 = 1, c1 = r
Similarly x = p'y + q'
⇒ y = `(x - "q'")/"p'"`
And z = r'y + s'
⇒ y = `(z - "s'")/"r'"`
∴ The equation becomes `(x - "q'")/"p'" = y/1 = (z - "s'")/"r'"` in which a2 = p', b2 = 1, c2 = r'
If the lines are perpendicular to each other, then
a1a2 + b1b2 + c1c2 = 0
pp' + 1.1 + rr' = 0
Hence, pp' + rr' + 1 = 0 is the required condition.
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