Question

Show that the lines x² − 4xy + y² = 0 and x + y = 10 contain the sides of an equilateral triangle. Find the area of the triangle. 

Answer 1

We find the joint equation of the pair of lines OA and OB through origin, each making an angle of 60° with x + y = 10 whose slope is −1.

Let OA(or OB) has slope m.

∴ its equation is y − mx        ...(1)

Also, tan 60° = `|("m" − (−1))/(1 + "m"(−1))|`

`therefore sqrt3 = |("m + 1")/(1 - "m")|`

Squaring both sides, we get,

`3 = ("m" + 1)^2/(1 - "m")^2`

∴ 3(1 − 2m + m²) = m² + 2m + 1

∴ 3 − 6m + 3m² = m² + 2m + 1

∴ 2m² − 8m + 2 = 0

∴ m² − 4m + 1 = 0

∴ `("y"/"x") − 4("y"/"x") + 1 = 0`   ...[By(1)]

∴ y² − 4xy + x² = 0

∴ x² − 4xy + y² = 0 is the joint equation of the two lines through the origin each making an angle of 60° with x + y = 10

∴ x² − 4xy + y² = 0 and x + y = 10 form a triangle OAB which is equilateral.

Let seg OM perpendicular line AB whose question is x + y = 10

∴ OM = `|(-10)/sqrt(1 + 1)| = 5sqrt2`

∴ area of equilateral Δ OAB `= ("OM")^2/sqrt3 = (5sqrt2)^2/sqrt3`

`= 50/sqrt3` sq units.