If the straight line $2x + 3y - 1 = 0$, $x + 2y - 1 = 0$ and $ax + by - 1 = 0$ form a triangle with origin as orthocentre, then $(a,b)$ is equal to:

Question

Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
If $B(\alpha, \beta)$, with $\beta < 4$, lies on $C$ such that the length of the arc $AB$ is $\frac{1}{6}$ of the perimeter of $C$, then $\beta - \sqrt{3}\alpha$ is equal to:

Answer 1

Point $ A $ is the intersection of lines $ AB $ and $ AC $. The equation of the line passing through $ A $ is:
\[ (x + 2y - 1) + \lambda (2x + 3y - 1) = 0 \] This line passes through the orthocenter $ (0,0) $. Substituting $ (0,0) $:
\[ -1 + \lambda (-1) = 0 \] \[ -1 - \lambda = 0 \] \[ \lambda = -1 \] Substituting $ \lambda = -1 $ into the equation yields:
\[ x + y = 0 \] Therefore, the equation of $ AD $ is:
\[ x + y = 0 \] Since $ AD \perp BC $:
\[ \frac{-1 - x}{a} = \frac{d}{b} = -1 \] This simplifies to:
\[ a + b = 0 \] Similarly, applying the condition $ BE \perp CA $ yields:
\[ a + 2b = 8 \] Solving the system of equations (3) and (4):
\[ a + b = 0 \] \[ a + 2b = 8 \] Subtracting Eq. (3) from Eq. (4):
\[ (a + 2b) - (a + b) = 8 - 0 \] \[ b = 8 \] Substituting $ b = 8 $ into Eq. (3):
\[ a + 8 = 0 \] \[ a = -8 \] The values are:
\[ a = -8, \quad b = 8 \]

If the straight line $2x + 3y - 1 = 0$, $x + 2y - 1 = 0$ and $ax + by - 1 = 0$ form a triangle with origin as orthocentre, then $(a,b)$ is equal to:

Show Hint

The Correct Option is C

Solution and Explanation

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