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# If the straight lines $2x + 3y - 1 = 0,x + 2y - 1 = 0,$ and $ax + by - 1 = 0$ form a triangle with the origin as orthocenter, then (a, b) is given byA.$\left( {6,4} \right)$ B.$\left( { - 3,3} \right)$ C.$\left( { - 8,8} \right)$ D.$\left( {0,7} \right)$

Hint: Orthocenter is the intersection point of the altitudes drawn from the vertices of the triangle to the opposite sides.
The general equation of the family of lines through the point of intersection of two given lines L1 and L2 is given by
${L_1} + \lambda {L_2} = 0$, where $\lambda$ is a parameter.

Formula Used:
Equation of a line passing through two points $\left( {{x_1},{y_1}} \right)\& \left( {{x_2},{y_2}} \right)$ is given as
$y - {y_1} = \left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)\left( {x - {x_1}} \right)$
Two lines L1 and L2 are perpendicular to each other if ${m_1}{m_2} = - 1$

Complete step by step solution:
According to the question,
Lines $2x + 3y - 1 = 0,x + 2y - 1 = 0$ and $ax + by - 1 = 0$ form a triangle, so to find the intersection point of lines $2x + 3y - 1 = 0,x + 2y - 1 = 0$ we have to convert the given equation into slope intercept form i.e. $y = mx + c$
Now, the equations are as follows
$\begin{gathered} 3y = 1 - 2x \ldots \left( 1 \right) \\ 2y = 1 - x \\ x = 1 - 2y \ldots \left( 2 \right) \\ \end{gathered}$
Substitute the value of x in equation (1)
$\begin{gathered} 3y = 1 - 2\left( {1 - 2y} \right) \\ 3y = 1 - 2 + 4y \\ y = 1 \\ \end{gathered}$

Put the value of y in equation (2)
$\begin{gathered} x = 1 - 2y \\ x = 1 - 2\left( 1 \right) \\ x = - 1 \\ \end{gathered}$
Hence, the intersection point is (-1, 1)
Line passing through (-1, 1) and origin is perpendicular to $ax + by - 1 = 0$
Required line ${L_1}:y = - x$ and ${L_2}:ax + by - 1 = 0$ are perpendicular as the product of their slope is -1.
By using ${m_1}{m_2} = - 1$, we get $a = - b \ldots \left( i \right)$
Also line passing through $x + 2y - 1 = 0$ and $ax + by - 1 = 0$, is
$x + 2y - 1 + \lambda \left( {ax + by - 1} \right) = 0$
It passes through origin, so
$\lambda = - 1$
Line is $x\left( {1 - a} \right) + y\left( {2 - b} \right) = 0$
This is perpendicular to line $2x + 3y - 1 = 0$
So, using ${m_1}{m_2} = - 1$, we get
$2a + 3b = 8 \ldots \left( {ii} \right)$
Then on solving equation (i) and (ii), we get
$\begin{gathered} a = 8 \\ b = - 8 \\ \end{gathered}$
Hence, (a, b) is (8, -8)
Option (C) is correct

Note: Position of orthocenter:
1)For an acute triangle, it lies inside the triangle.
2)For an obtuse triangle, it lies outside of the triangle.
3)For a right-angled triangle, it lies on the vertex of right angle