points C,D, and E are collinear on CE, and CD:DE = 3/5. C is located at (1,8), D is located at (4,5), and E is located at (x,y). What are the values of x and y

Answer:The point E is located at (9,0)x=9, y=0Step-by-step explanation:we have thatPoints C,D, and E are collinear on CEPoint D is between point C and point Ewe know that[tex]CE=CD+DE[/tex] -----> equation A (by addition segment postulate)[tex]\frac{CD}{DE}=\frac{3}{5}[/tex][tex]CD=\frac{3}{5}DE[/tex] ------> equation Bthe formula to calculate the distance between two points is equal to[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]Find the distance CD   we haveC(1,8), D(4,5)substitute in the formula[tex]CD=\sqrt{(5-8)^{2}+(4-1)^{2}}[/tex][tex]CD=\sqrt{(-3)^{2}+(3)^{2}}[/tex][tex]CD=\sqrt{18}\ units[/tex]Find the distance DEsubstitute the value of CD in the equation B and solve for DE[tex]\sqrt{18}=\frac{3}{5}DE[/tex][tex]DE=\frac{5\sqrt{18}}{3}\ units[/tex]Find the distance CE[tex]CE=CD+DE[/tex]we have[tex]DE=\frac{5\sqrt{18}}{3}\ units[/tex][tex]CD=\sqrt{18}\ units[/tex]substitute the values in the equation A[tex]CE=\sqrt{18}+\frac{5\sqrt{18}}{3}[/tex][tex]CE=\frac{8\sqrt{18}}{3}[/tex]Applying the formula of distance CEwe have[tex]CE=\frac{8\sqrt{18}}{3}[/tex]C(1,8), E(x,y)    substitute in the formula of distance[tex]\frac{8\sqrt{18}}{3}=\sqrt{(y-8)^{2}+(x-1)^{2}}[/tex]squared both sides[tex]128=(y-8)^{2}+(x-1)^{2}[/tex]  -----> equation CApplying the formula of distance DEwe have[tex]DE=\frac{5\sqrt{18}}{3}\ units[/tex]D(4,5), E(x,y)    substitute in the formula of distance[tex]\frac{5\sqrt{18}}{3}=\sqrt{(y-5)^{2}+(x-4)^{2}}[/tex]squared both sides[tex]50=(y-5)^{2}+(x-4)^{2}[/tex]  -----> equation Dwe have the system[tex]128=(y-8)^{2}+(x-1)^{2}[/tex]  -----> equation C[tex]50=(y-5)^{2}+(x-4)^{2}[/tex]  -----> equation DSolve the system by graphingThe intersection point both graphs is the solution of the systemThe solution is the point (9,0)thereforeThe point E is located at (9,0)see the attached figure to better understand the problem