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प्रश्न
A ∆OPQ is formed by the pair of straight lines x2 – 4xy + y2 = 0 and the line PQ. The equation of PQ is x + y – 2 = 0, Find the equation of the median of the triangle ∆ OPQ drawn from the origin O
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उत्तर
The equation of the given pair of lines is
x2 – 4xy + y2 = 0 .......(1)
The equation of the line PQ is
x + y – 2 = 0
y = 2 – x .......(2)
To find the coordinates of P and Q, .
Solve equations (1) and (2)
(1) ⇒ x2 – 4x (2 – x) + (2 – x)2 = 0
x2 – 8x + 4x2 + 4 – 4x + x2 = 0
6x2 – 12x + 4 = 0
3x2 – 6x + 2 = 0
x = `(6 +- sqrt(6^2 - 4 xx 3 xx 2))/(2 xx 3)`
= `(6 +- sqrt(36 - 24))/6`
= `(6 +- sqrt(12))/6`
= `(6 +- 2sqrt(3))/6`
= `(3 +- sqrt(3))/3`
When x = `(3 +- sqrt(3))/3`, y = `2 - (3 +- sqrt(3))/3`
y = `(6 - 3 - sqrt(3))/3`
= `(3 - sqrt(3))/3`
When x = `(3 - sqrt(3))/3`, y = `2 - (3 - sqrt(3))/3`
y = `(6 - 3 + sqrt(3))/3`
= `(3 + sqrt(3))/3`
∴ P is `((3 + sqrt(3))/3, (3 - sqrt(3))/3)`
and
Q is `((3 - sqrt(3))/3, (3 + sqrt(3))/3)`
The midpoint of PQ is
D = `(((3 + sqrt(3))/3 + (3 - sqrt(3))/3)/3, ((3 - sqrt(3))/3 + (3 + sqrt(3))/3)/3)`
= `((3 + sqrt(3) + 3 - sqrt(3))/6, (3 - sqrt(3) + 3 + sqrt(3))/6)`
= `(6/6, 6/6)`
= (1, 1)
The equation of the median drawn from 0 is the equation of the line joining 0(0, 0) and D(1, 1)
`(x - 0)/(1 - 0) = (y - 0)/(1 - 0)`
⇒ `x/1 = y/1`
∴ The required equation is x = y
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