In the diagram, the areas of adc and dcb are in a ratio of 3: 4. what are the coordinates of point c

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)

I have remind and 0+2=0

Step-by-step explanation:

:)