`DeltaOAB` is formed by the lines `x^2-4xy +y^2= 0` and the line AB. The equation of line AB is `2x + 3y-1 = 0`.Find the equation of the median of the triangle drawn from the origin.

Explanation

Let D be the midpoint of seg AB where A is `(x_(1), y_(1))` and B is `(x_(2), y_(2))`
Then D has coordinates `((x_(1)+x_(2))/(2),(y_(1)+y_(2))/(2))`
The joint (combined) equation of the lines OA and OB is `x^(2) - 4xy + y^(2) = 0` and the equation of the line AB is `x + y -2 = 0`.
`therefore` points A and B satisfy the equations `x + y - 2 = 0` and `x^(2) - 4xy + y^(2) = 0` simultaneously.
We eliminate x from the above equations.
Put x = 2 - y in the equation `x^(2) - 4xy + y^(2) = 0`, we get,
`(2-y)^(2) - 4(2-y) y + y^(2) = 0`
`therefore 4-4y + y^(2) - 8y + 4y^(2) + y^(2) = 0`
`therefore 6y^(2) - 12y + 4 = 0 " " therefore3y^(2) - 6y + 2 = 0`
The roots `y_(1) and y_(2)` of the above quadratic equation are the y-coordinates of the point A and B.
`therefore y_(1) + y_(2) = (6)/(3) = 2`
`therefore` y-coordinate of D ` = (y_(1) + y_(2))/(2) = 1.`
Since D lies on the line AB, we can find the x-coordinate of D as
`x + 1 -2 = 0 " " therefore x = 1`
`therefore` D is (1, 1)
`therefore` equation of the median AD is `(y-0)/(x-0) = (1-0)/(1-0), i..e, y = x, i.e., x - y = 0.`