Question

Sol. [Key Concept: Lines through $(2,3)$ with axis-triangle area $12$ ⇒ a quadratic in the slope; its roots count the lines.]

Equation of any line through ( 2,3 ) is $y - 3 = m(x - 2)$

$$y = mx - 2m + 3 $$

with the help of the fig. area of $\Delta OAB = 12$

i.e., $\frac{1}{2}\left( \frac{2m - 3}{m} \right)(3 - 2m) = \pm 12$

taking + sign we get $(2m + 3)^{2} = 0$

this gives one value of $m = - 3/2$

taking negative sign we get

$$4m^{2} - 36m + 9 = 0(D > 0) $$quadratic in $m$ gives 2 values of $m$

$\Rightarrow 3$ st. lines are possible.

Key insight: the area condition gives a quadratic in the slope with a positive discriminant plus one degenerate case ⇒ three such lines.