Q:

in the diagram, the areas of ADC and DCB are in a ratio of 3:4. what are the coordinates of point C

Accepted Solution

A:
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)