\(\because\) It is clear that points (0, y/b) and (4/3, 0) on line 3x + by = 4, will form a right triangle with A(0, 0) and B(0, 4).
Let (x, y) be other point on line 3x + by = 4, which from a right triangle with A(0, 0) and B(0, 4).
\(\therefore\) (Slope of line formed by points (x, y) and (0, 4)) = -1
(\(\because\) product of slops of two perpendicular lines is -1).
\(\therefore\) (\((\frac{y-0}{x-0})\times(\frac{4-y}{0-x})\) = -1
⇒ y(4 - y) = x2
⇒ 4y - y2 = (\(\frac{4-by}3\))2
(\(\because\) point (x, y) will lie on line 3x + by = 4)
⇒ 9(4y - y2) = b2y2 - 8by + 16
⇒ (b2 + 9)y2 + (36 - 8b)y + 16 = 0
\(\because\) It shows a unique point
\(\therefore\) (36 - 8b)2 = 4(b2 + 9) x 16
⇒ 362 + 64b2 - 36 x 16b = 64b2 + 9 x 64
⇒ 36 x 166 = 362 - 9 x 64 = 62(62 - 24)
⇒ b = \(\frac{36(36-16)}{36\times16}=\frac{20}{16} = \frac54\)
Hence, value of b is 5/4 for which three distinct points on line 3x + by = 4 will form distinct points on line 3x + by = 4 will form a right triangle with point A(0, 0) and B(0, 4)