Question

Find the centre and radius of the circle which is inscribed in the triangle formed by the straight lines whose equations are:
$y=0$, $12x-5y=0$, and $3x+4y-7=0$.

Answer 1

$y=\mathrm{0..................}\left(i\right)12x-5y=\mathrm{0............}\left(ii\right)3x+4y-7=\mathrm{0.............}\left(iii\right)$

Solving $\left(ii\right)$ and $\left(iii\right)$ we get $A(\frac{5}{9},\frac{4}{3})$

Solving $\left(iii\right)$ and $\left(i\right)$ we get $B(\frac{7}{3},0)$

solving $\left(i\right)$ and $\left(ii\right)$ we get $C(0,0)$

$BC=a=\sqrt{{(\frac{7}{3}-0)}^2+{(0-0)}^2}=\frac{7}{3}CA=b=\sqrt{{(\frac{5}{9}-0)}^2+{(\frac{4}{3}-0)}^2}=\frac{13}{9}AB=c=\sqrt{{(\frac{5}{9}-\frac{7}{3})}^2+{(\frac{4}{3}-0)}^2}=\frac{20}{9}$

Coordinates of centre are

$[\left\{\frac{\frac{7}{3}\times \frac{5}{9}+\frac{13}{9}\times \frac{7}{3}+\frac{20}{9}\times 0}{\frac{7}{3}+\frac{13}{9}+\frac{20}{9}}\right\},\left\{\frac{\frac{7}{3}\times \frac{4}{3}+\frac{13}{9}\times 0+\frac{20}{9}\times 0}{\frac{7}{3}+\frac{13}{9}+\frac{20}{9}}\right\}](\frac{7}{9},\frac{14}{27})$

Radius is equal to distance from $y=0$

$r=\frac{14}{27}$