Finding Orthocenter in Coordinate Geometry

Question

If a triangle is formed by the equations \begin{gather}2x+3y-1=0\\ ~~x+2y-1=0\\ ax+by-1=0\end{gather} and has its orthocentre at origin, then what are the values of $a$ and $b$? (Please also tell me the Idea and steps involved in the solution)

Answer 1

HINT:

From the definition of Orthocenter,

the perpendicular of $2x+3y-1=0\ \ \ \ (1)$ through $(0,0)$ is $3x-2y=0\ \ \ \ (2)$

Now $ax+by-1=0,(2)$ and $x+2y-1=0$ are concurrent

Apply the same method for the perpendicular of $x+2y-1=0$

Ultimately, we have two linear equations in $a,b$

Answer 2

The figure drawn is not in scale. In which, we let the equations be:-

$(A): 2x + 3y – 1 = 0$

$(B): x + 2y – 1 = 0$

$(C): ax + by – 1 = 0$

Step-1 (A) and (B) intersect at … P(–1, 1), which is one of the vertices of ⊿PQR.

Step-2 The equation of the line passing through O and perpendicular to PR is … (D): 2x – y = 0.

Step-3 Solving (D) and (A) gives $Q = … = (\frac {1}{8}, \frac {2}{8})$.

Step-4 Slope of OP = … = –1 and therefore slope of (C) is … 1.

Step-5 Therefore, we let the equation of (C) be y = x + k for some k.

Step-6 Putting Q in (C), we get the actual equation of (C), which is $y = x + (\frac {1}{8})$

Step-7 Re-writing the above equation to get the values of a and b.