Question

Find the coordinates of the incentre and centroid of the triangle whose sides have the equations $3x-4y=0,12y+5x=0$ and $y-15=0.$

Answer 1

Let ABC be the triangle whose sides BC, CA and AB have the equations

$y-15=0,BC$

$3x-4y=0,AC$

$5x+12y=0AB$

Solving these equations pair wise we can obtain the coordinates of the vertices A, B, C as

$A(0,0)B(-36,15)C(20,15)$ respectively

Centroid $(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})(\frac{-36+20+0}{3},\frac{15+15+0}{3})=(\frac{-16}{3},10)$

For incentre, we have

$a=BC=\sqrt{{56}^2+0}=56$

$b=CA=\sqrt{{20}^2+{15}^2}=25$

$c=AB=\sqrt{{36}^2+{16}^2}=39$

Coordinates of incentre are $(\frac{56\times 0+25x-36+39\times 20}{36+25+39},\frac{56\times 0+25\times 15+39\times 15}{36+25+39})$

$=(-1,8)$