Question

ΔOAB is formed by the lines x² - 4xy + y² = 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 O.

Answer 1

Let D be the midpoint of seg AB where A is (x₁, y₁) and B is (x₂, y₂).

Then D has coordinates ${\displaystyle (\frac{{\text{x}}_1+{\text{x}}_2}{2},\frac{{\text{y}}_1+{\text{y}}_2}{2})}$.

The joint (combined) equation of the lines OA and OB is x² - 4xy + y² = 0 and the equation of the line AB is 2x + 3y - 1 = 0

∴ points A and B satisfy the equations 2x + 3y - 1 = 0 and x² - 4xy + y² = 0 simultaneously. 

We eliminate x from the above equations, i.e., put x = ${\displaystyle \frac{1-\text{3y}}{2}}$ in the equation x² - 4xy + y² = 0, we get, 

∴ ${\displaystyle {\left(\frac{1-3\text{y}}{2}\right)}^2-4\left(\frac{1-\text{3y}}{2}\right)\text{y}+{\text{y}}^2=0}$

∴ ${\displaystyle {(1-\text{3y})}^2-8(1-\text{3y})\text{y}+{\text{4y}}^2=0}$

∴ 1 - 6y + 9y² - 8y + 24y² + 4y² = 0 

∴ 37y² - 14y + 1 = 0

The roots y₁ and y₂ of the above quadratic equation are the y-coordinates of the points A and B.

∴ y₁ + y₂ = ${\displaystyle -\frac{\text{b}}{\text{a}}=\frac{14}{37}}$

∴ y-coordinate of D = ${\displaystyle \frac{{\text{y}}_1+{\text{y}}_2}{2}=\frac{7}{37}}$

Since D lies on the line AB, we can find the x-coordinate of D as

2x + 3${\displaystyle \left(\frac{7}{37}\right)-1=0}$

∴ 2x = ${\displaystyle 1-\frac{21}{37}=\frac{16}{37}}$

∴ x = ${\displaystyle \frac{8}{37}}$

∴ D is ${\displaystyle (\frac{8}{37},\frac{7}{37})}$

∴ equation of the median OD is ${\displaystyle \frac{\text{x}}{8/37}=\frac{\text{y}}{7/37}}$

i.e. 7x - 8y = 0