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Question
Find the equations of the diagonals of the parallelogram PQRS whose vertices are P(4, 2, – 6), Q(5, – 3, 1), R(12, 4, 5) and S(11, 9, – 2). Use these equations to find the point of intersection of diagonals.
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Solution
Equation of diagonal PR is i.e., P(4, 2, – 6) and R(12, 4, 5) is
`(x - 4)/(12 - 4) = (y - 2)/(4 - 2) = (z + 6)/(5 + 6)`
`\implies (x - 4)/8 = (y - 2)/2 = (z + 6)/11` ...(i)
Equation of diagonal QS is i.e., Q(5, – 3, 1) and S(11, 9, – 2) is
`(x - 5)/(11 - 5) = (y + 3)/(9 + 3) = (z - 1)/(-2 - 1)`
`\implies (x - 5)/6 = (y + 3)/12 = (z - 1)/-3` ...(ii)
For finding point of intersection.
Let `(x - 4)/8 = (y - 2)/2 = (z + 6)/11` = λ
x = 8λ + 4, y = 2λ + 2, z = 11λ – 6
Since, the lines (i) and (ii) are intersecting, the point (x, y, z) must satisfy the second line.
∴ `(8λ + 4 - 5)/6 = (2λ + 2 + 3)/12 = (11λ - 6 - 1)/-3`
`\implies (8λ - 1)/6 = (2λ + 5)/12 = (11λ - 7)/-3`
`\implies (8λ - 1)/6 = (2λ + 5)/12`
`\implies` 16λ – 2 = 2λ + 5
`\implies` 14λ = 7
λ = `1/2`
∴ Point of intersection is
x = `8 xx 1/2 + 4` = 8
y = `2 xx 1/2 + 2` = 3
z = `11 xx 1/2 - 6` = `11/2 - 6` = `(-1)/2`
Thus, the diagonals intersect at `(8, 3, (-1)/2)`
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