Question

If $x_1,y_1$ are the roots of $x^2+8x-20=0$ and $x_2,y_2$ are the roots of $4x^2+32x-57=0$ and $x_3,y_3$ are the roots of $9x^2+72x-112=0$ such that $y_i<0,$ 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

Answer 1

The correct option is A are collinear

$x^2+8x-20=0$
$\to (x+10)(x-2)=0$

$x=-10,x=2$

Therefore

$(x_1,y_1)=(2,-10)$...(i)

Similarly

$4x^2+32x-57=0$

$(x-\frac{3}{2})(x+\frac{19}{2})=0$

$x=\frac{3}{2},\frac{-19}{2}$

Hence

$(x_2,y_2)=(\frac{3}{2},\frac{-19}{2})$...(ii)

$9x^2+72x-112=0$

$(x-\frac{4}{3})(x+\frac{28}{3})=0$

$x=\frac{4}{3},\frac{-28}{3}$

Hence

$(x_3,y_3)=(\frac{4}{3},\frac{-28}{3})$...(iii)

Therefore the three points are

$A=(x_1,y_1)=(2-10)$

$B=(x_2,y_2)=(\frac{3}{2},\frac{-19}{2})$

$C=(x_3,y_3)=(\frac{4}{3},\frac{-28}{3})$

Thus, there are two possibilities

I) The points are co-linear.

II) The points are non-co-linear (forming a triangle).

Hence, first we check for co-linearity.

If the points are co-linear, then the must lie on a single straight line.

The equation of the line passing through $A=(2,-10)$ and $B=(\frac{3}{2},\frac{-19}{2})$ is

$\frac{y+10}{x-2}=\frac{y+\frac{19}{2}}{x-\frac{3}{2}}$

$\to x+y=-8$.

Now if C is collinear with A and B, then it must satisfy the above equation of the straight line.

Substituting $C=(x_3,y_3)$ in the given line gives us

$\frac{4}{3}-\frac{28}{3}$

$=\frac{-24}{3}$

$=-8$

$=RHS$

Hence the points are collinear.