If x1, y1 are roots of x2+8 x-20=0, x2, y2 are the roots of 4 x2 +32 x-57=0 and x3, y3 are the roots of 9 x2+72 x-112=0 then the points ( x 1, y 1),( x 2, y 2) and ( x 3, y 3)

Question

If x1, y1 are roots of x2+8 x-20=0, x2, y2 are the roots of 4 x2 +32 x-57=0 and x3, y3 are the roots of 9 x2+72 x-112=0 then the points ( x 1, y 1),( x 2, y 2) and ( x 3, y 3) (A) are collinear (B) form an equilateral triangle (C) form a right angled isosceles triangle (D) are concyclic

Tardigrade Admin · Accepted Answer

x1+y1=-8, x2+y2=-(32/4)=-8, x3+y3=-(72/9)=-8 ∵ Area of the triangle formed by ( x 1, y 1),( x 2, y 2) and ( x 3, y 3) is 0 . Hence, the points ( x 1, y 1),( x 2, y 2) and ( x 3, y 3) are collinear.

Q. If $x_{1}, y_{1}$ are roots of $x^{2} + 8 x - 20 = 0, x_{2}, y_{2}$ are the roots of $4 x^{2}$ $+ 32 x - 57 = 0$ and $x_{3}, y_{3}$ are the roots of $9 x^{2} + 72 x - 112 = 0$ then the points $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$

are collinear

form an equilateral triangle

form a right angled isosceles triangle

are concyclic

$x_{1} + y_{1} = - 8$ ,
$x_{2} + y_{2} = - \frac{32}{4} = - 8$ ,
$x_{3} + y_{3} = - \frac{72}{9} = - 8$
$∵$ Area of the triangle formed by
$(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is $0.$
Hence, the points $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ are collinear.

Q. If x1​,y1​ are roots of x2+8x−20=0,x2​,y2​ are the roots of 4x2 +32x−57=0 and x3​,y3​ are the roots of 9x2+72x−112=0 then the points (x1​,y1​),(x2​,y2​) and (x3​,y3​)