Show that the lines `x^(2) - 4xy + y^(2) = 0` and `x + y = 10` contain the sides of an equilateral triangle.

Explanation

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.
`therefore` its equation is `y = mx" "….(1)`
Also, `60^(@) = |(m-(-1))/(1+m(-1))|`
`therefore sqrt(3) = |(m+1)/(1-m)|`
Squaring both sides, we get
` 3= ((m+1)^(2))/((1-m)^(2))`
`therefore 3(1-2m+m^(2))=m^(2)+2m+1`
`therefore 3-6m+3m^(2)=m^(2)+2m+1`
`therefore 2m^(2)-8m +2 = 0`
`therefore m^(2) - 4m + 1 = 0`
`therefore((y)/(x))^(2)-4((y)/(x))+1 =0" "`...[By(1)]
`therefore y^(2)-4xy +x^(2) =0`
`therefore x^(2) - 4xy + y^(2) = 0` is the joint equation of the two lines through the origin each making an angle of `60^(@)` with x + y = 10
`therefore x^(2) - 4xy + y^(2) = 0` and `x + y = 10` form a triangle OAB which is equilateral.