Question

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

Answer 1

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