the complete question in the attached figure
we know
that
dAB=dAC+dCB
dAC=√((8-1)²+(-2+9)²)=√98
Area triangle ADC=dAC*CD/2
Area triangle DCB=dCB*CD/2
Area triangle ADC/Area triangle DCB=3/4
[dAC*CD/2]/[dCB*CD/2]=3/4
dAC=(3/4)*dCB-----------------------> equation (1)
dAB=dAC+dCB=√98
dAC=√98-dCB------------------------ > equation (2)
(1)=(2)
(3/4)*dCB=√98-dCB------------------>
(7/4)*dCB=√98
dCB=(4/7)√98
dAC=√98-dCB--------
> √98-(4/7)√98-----à (3/7) )√98
dAC=(3/7) )√98
find the slope point A (1.-9) and point B (8,-2)
m=(-2+9)/(8-1)=7/7=1-------------- > 45°
dAC=(3/7)√98
the component x of dAC is
dACx=(3/7)√98*cos45°=(3/7)√98*√2/2=(3/7)√196=3
the component y of dAC is
dACy=dACx=3
the coordinates of the point C are
point A(1.-9)
Cx=Ax+dACx----------> 1+3=4
Cy=Ay+dACy----------> -9+3=-6
the coordinates C(4,-6)
the answer is C(4,-6)
