ction point by combining their equations:

Equation of the first line: $x - 2y + 3 = 0$

Equation of the second line: $2x - y - 1 = 0$

The general equation of a line passing through their intersection is a linear combination of the two lines:

$ x - 2y + 3 + \lambda(2x - y - 1) = 0 $

Simplifying:

$ x(1 + 2\lambda) + y(-2 - \lambda) + (3 - \lambda) = 0 $

This can be rearranged into intercept form:

$ \frac{x}{\frac{\lambda-3}{1+2\lambda}} + \frac{y}{\frac{3-\lambda}{\lambda+2}} = 1 $

This gives the coordinates of points $A$ and $B$ as:

$A\left(\frac{\lambda-3}{1+2\lambda}, 0\right)$

$B\left(0, \frac{3-\lambda}{\lambda+2}\right)$

To find the point dividing $AB$ in the ratio $-2:3$, apply the section formula:

For the x-coordinate $h$:

$ h = \frac{3 \left(\frac{\lambda-3}{1+2\lambda}\right)}{-2 + 3} = 3 \left(\frac{\lambda-3}{1+2\lambda}\right) $

For the y-coordinate $k$:

$ k = \frac{-2 \left(\frac{3-\lambda}{\lambda+2}\right)}{-2 + 3} = 2 \left(\frac{3-\lambda}{\lambda+2}\right) $

Now, solving for $\lambda$ using the x-coordinate $h$:

$ h + 2h\lambda = 3\lambda - 9 \Rightarrow 2h\lambda - 3\lambda = -h - 9 $

Solving for $\lambda$:

$ \lambda = \frac{-h-9}{2h-3} $

Substituting into the expression for $k$:

$ k = 2 \left[\frac{\frac{-h-9}{2h-3} - 3}{\frac{-h-9}{2h-3} + 2}\right] = \frac{-14h}{3h-15} $

Thus, the equation of the locus is:

$ 3hk - 15k + 14h = 0 $

Therefore, the required equation is:

$ 14x + 3xy - 15y = 0 $

Straight Lines And Pair Of Straight Lines

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