The in-radius of the triangle formed by the axes and the line 4x + 3y – 12 = 0 is:

$r = 2$ $r=\frac{1}{2}$ $r = 1$ $r=\frac{1}{4}$

$r = 1$

$\frac{x}{3}+\frac{y}{4}=1$ ⇒ The sides are 3, 4, 5  $⇒\sqrt{a^2+b^2}=5$ $area\, Δ =\frac{1}{2}.3.4=6⇒r=\frac{Δ}{s}=\frac{6}{6}=1$