The radius of the incircle of the triangle formed by the x-axisand the lines 3x + 4y - 24 = 0, 3x - 4y + 24 = 0 is

For 3x+4y-24=0 can be re-written as $$3x+4y=24$$

=>$$\frac{3x+4y\ \ }{24}=1$$

=>$$\ \frac{\ x}{8}+\ \frac{\ y}{6}=1$$.

So, this becomes the equation of the straight line of the form $$\ \frac{\ x}{a}+\ \frac{\ y}{b}=1$$, where a and b are x and y intercepts respectively. The X and Y intercepts for this equation are 8 and 6 respectively.

Similarly 3x-4y+24=0 can be re-written as $$\ \frac{\ x}{-8}+\ \frac{\ y}{6}=1$$. The X and Y intercepts for this equation are 8 and -6 respectively.

So, we get the following triangle and the incircle can be drawn as:

So, the triangle has sides of length 10,10 and 16.

Area of the triangle= $$\ \frac{\ 1}{2}\times\ Base\times\ Height$$= $$\ \frac{\ 1}{2}\times\ 16\times\ 6$$= 48

We know that r*s= Area of the triangle, where r is the inradius and s is the semi-perimeter. s= $$\ \frac{\ 1}{2}\left(16+10+10\right)$$= 18

So, 18r= 48

=> Or, r= $$\ \frac{\ 48}{18}=\ \ \frac{\ 8}{3}$$