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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.
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Solution
Given x = py + q ⇒ y = `(x - q)/p`
And z = ry + s ⇒ y = `(z - s)/r`
So, 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"'")`
Hence, 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
Thus, the given lines are perpendicular if pp' + rr' + 1 = 0.
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