Question

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:

• A. (6,4)
• B. (-3,3)
• C. (-8,8)
• D. (0,7)

Hint:

Understanding orthocentre properties helps simplify the problem-solving approach.

Answer 1

The Correct Option is C

Here, point \( A \) is the intersection of line \( AB \) and \( AC \). So, the equation of the line passing through \( A \) is given by:
\[ (x + 2y - 1) + \lambda (2x + 3y - 1) = 0 \] This line passes through the orthocenter \( (0,0) \), hence substituting \( (0,0) \):
\[ -1 + \lambda (-1) = 0 \] \[ -1 - \lambda = 0 \] \[ \lambda = -1 \] Substituting \( \lambda = -1 \) in the equation above, we get:
\[ x + y = 0 \] Thus, the equation of \( AD \) is:
\[ x + y = 0 \] Since \( AD \perp BC \), therefore:
\[ \frac{-1 - x}{a} = \frac{d}{b} = -1 \] which simplifies to:
\[ a + b = 0 \] Similarly, by applying the condition that \( BE \) is perpendicular to \( CA \), we obtain:
\[ a + 2b = 8 \] Now, 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 \( b = 8 \) in Eq. (3):
\[ a + 8 = 0 \] \[ a = -8 \] Thus, the values are:
\[ a = -8, \quad b = 8 \]