Question

Co-ordinates of the incentre of the triangle formed by the lines $3x+4y=12$ and the co-ordinate axes is

Answer 1

The co-ordinate axes has its origin at $O(0,0)$ and triangle is formed by the line $3x+4y=12$

The line $3x+4y=12$ cuts the $x-$axis at $y=0$ $\therefore$ co-ordinate of $x-$axis is $(4,0)$ by substituting $y=0$ in $3x+4y=12$ then $3x=12$ or $x=4$

The line $3x+4y=12$ cuts the $y-$axis at $x=0$ $\therefore$ co-ordinate of $y-$axis is $(0,3)$ by substituting $y=0$ in $3x+4y=12$ then $4y=12$ or $y=3$

Now the distance $OA=4$units and that of $OB=3$units

Using formula for incentre of a triangle we have

$I(x,y)=(\frac{a{x}_1+b{x}_2+c{x}_3}{3},\frac{a{y}_1+b{y}_2+c{y}_3}{3})$

Since $△BOA$ is a right angled triangle

Using Pythagoras theorem, ${AB}^2={OA}^2+{OB}^2=4^2+3^2=16+9=25$

or $AB=5$ sq.units

$\therefore$ $b=4,a=3,c=5$

$I(x,y)=(\frac{3\times 4+0+0}{3+4+5},\frac{0+4\times 3+0}{3+4+5})=(\frac{12}{12},\frac{12}{12})=(1,1)$