Straight LineHardBloom L3

Question

Let $x_{1}$ and $y_{1}$ be the roots of $x^{2} + 8x - 2009 = 0;x_{2}$ and $y_{2}$ be the roots of $3x^{2} + 24x - 2010 = 0$ and $x_{3}$ and $y_{3}$ be the roots of $9x^{2} + 72x - 2011 = 0$. The points $A\left( x_{1},y_{2} \right)B\left( x_{2},y_{2} \right)$ and $C\left( x_{3},y_{3} \right)$ :

Options

A.can not lie on a circle
B.form a triangle of area 2 sq. units
C.form a right-angled triangle
D.are collinear

Solution

Sol. [Key Concept: The three points share a common $x:y$ relation, so they are collinear (and hence cannot lie on a circle).]

$$D = \left| \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} \right|$$

Applying $C_{1} \rightarrow C_{1} + C_{2}$, we get,

$$D = \left| \begin{matrix} x_{1} + y_{1} & y_{1} & 1 \\ x_{2} + y_{2} & y_{2} & 1 \\ x_{3} + y_{3} & y_{3} & 1 \end{matrix} \right| = \left| \begin{matrix} - 8 & y_{1} & 1 \\ - 8 & y_{2} & 1 \\ - 8 & y_{3} & 1 \end{matrix} \right| = 0$$

✔ (a) collinear points cannot lie on a circle.

✘ (b) area is 0, not 2.

✘ (c) a degenerate (collinear) set is not a right triangle.

✔ (d) the points are collinear.

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