Question

Find the centre and radius of the circle which is inscribed in the triangle formed by the straight lines whose equations are
$3x+4y+2=0$, $3x-4y+12=0$, and $4x-3y=0$.

Answer 1

$3x+4y+2=\mathrm{0...................}\left(i\right)3x-4y+12=\mathrm{0...................}\left(ii\right)4x-3y=\mathrm{0..........................}\left(iii\right)$

Solving $\left(ii\right)$ and $\left(iii\right)$ we get $A(\frac{36}{7},\frac{48}{7})$

Solving $\left(i\right)$ and $\left(iii\right)$ we get $B(-\frac{6}{25},-\frac{8}{25})$

Solving $\left(i\right)$ and $\left(ii\right)$ we get $C(-\frac{7}{3},\frac{5}{4})$

$BC=a=\sqrt{{(-\frac{6}{25}+\frac{7}{3})}^2+{(-\frac{8}{25}-\frac{5}{4})}^2}=\frac{157}{60}CA=b=\sqrt{{(\frac{36}{7}+\frac{7}{3})}^2+{(\frac{48}{7}-\frac{5}{4})}^2}=\frac{785}{84}AB=c=\sqrt{{(\frac{36}{7}+\frac{6}{25})}^2+{(\frac{48}{7}+\frac{8}{25})}^2}=\frac{314}{35}$

Coordinates of incentre are

$[\left\{\frac{\frac{157}{60}\times \frac{36}{7}+\frac{785}{84}\times \frac{-6}{25}+\frac{314}{35}\times \frac{-7}{3}}{\frac{157}{60}+\frac{785}{84}+\frac{314}{35}}\right\},\left\{\frac{\frac{157}{60}\times \frac{48}{7}+\frac{785}{84}\times \frac{-8}{25}+\frac{314}{35}\times \frac{5}{4}}{\frac{157}{60}+\frac{785}{84}+\frac{314}{35}}\right\}](-\frac{13}{28},\frac{35}{28})$

Radius is equal to perpendicular distance from $4x-3y=0$

$r=\frac{|4\times \frac{-13}{28}-3\times \frac{35}{28}|}{\sqrt{4^2+3^2}}=\frac{157}{140}$