Let the equation of line forming a triangle having area 12sq. units is ‌
x
a
+‌
y
b
=1 ...(i) So, |ab|=24...(ii) ∵ Line (i) passes through point (2,3), so ‌
2
a
+‌
3
b
=1...(iii) from Eqs. (ii) and (iii), we get |2b+3a|=24...(iv) If a and b are positive, then ab=24 and 3a+2b=24 ⇒‌‌a(‌
24−3a
2
)=24 3a2−24a+48=0 ∵ Discriminant =242−4×3×48=0 ∴‌‌a=4 and b=6 If a is positive and b is negative then ab=−24 and 3a+2b=24.
⇒‌‌2(‌
24−3a
2
)=−24⇒3a2−24a−48=0
∵ Discriminant =242+4×3×48>0 ∴‌‌a=‌
24±√2×242
6
=4+4√2‌‌∵a>0 and b=6−6√2 If a is negative and b is positive, then ab=−24 and 3a+2b=24 ⇒‌‌a=4−4√2 and b=6+6√2 And at the last it is not possible that a and b both are negative. So, 3 triangles are possible.