If the line 3x+4y−24=0 intersects the x-axis at the point A and the y-axis at the point B, then the incentre of the triangle OAB, where O is the origin, is:
Solutions · 1
Official solution
A=(8,0),B=(0,6),O=(0,0); OA=8,OB=6,AB=10.
Incentre =(a+b+ca⋅x1+b⋅x2+c⋅x3,a+b+ca⋅y1+b⋅y2+c⋅y3) with a=OB(opposite A)=6... using the standard labeling (side opposite each vertex):
\[I = \left(\frac{8(0)+6(8)+10(0)}{8+6+10},\frac{8(6)+6(0)+10(0)}{24}\right) = \left(\frac{48}{24},\frac{48}{24}\right)=(2,2)\]