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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 - Mathematics

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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

योग

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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Pair of Straight Lines
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Two Dimensional Analytical Geometry - Exercise 6.4 [पृष्ठ २८२]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 6 Two Dimensional Analytical Geometry
Exercise 6.4 | Q 10 | पृष्ठ २८२

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