Question

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 1

Answer:

The point E is located at (9,0)

x=9, y=0

Step-by-step explanation:

we have that

Points C,D, and E are collinear on CE

Point D is between point C and point E

we know that

$CE=CD+DE$ -----> equation A (by addition segment postulate)

$\frac{CD}{DE}=\frac{3}{5}$

$CD=\frac{3}{5}DE$ ------> equation B

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Find the distance CD  

we have

C(1,8), D(4,5)

substitute in the formula

$CD=\sqrt{(5-8)^{2}+(4-1)^{2}}$

$CD=\sqrt{(-3)^{2}+(3)^{2}}$

$CD=\sqrt{18}\ units$

Find the distance DE

substitute the value of CD in the equation B and solve for DE

$\sqrt{18}=\frac{3}{5}DE$

$DE=\frac{5\sqrt{18}}{3}\ units$

Find the distance CE

$CE=CD+DE$

we have

$DE=\frac{5\sqrt{18}}{3}\ units$

$CD=\sqrt{18}\ units$

substitute the values in the equation A

$CE=\sqrt{18}+\frac{5\sqrt{18}}{3}$

$CE=\frac{8\sqrt{18}}{3}$

Applying the formula of distance CE

we have

$CE=\frac{8\sqrt{18}}{3}$

C(1,8), E(x,y)    

substitute in the formula of distance

$\frac{8\sqrt{18}}{3}=\sqrt{(y-8)^{2}+(x-1)^{2}}$

squared both sides

$128=(y-8)^{2}+(x-1)^{2}$  -----> equation C

Applying the formula of distance DE

we have

$DE=\frac{5\sqrt{18}}{3}\ units$

D(4,5), E(x,y)    

substitute in the formula of distance

$\frac{5\sqrt{18}}{3}=\sqrt{(y-5)^{2}+(x-4)^{2}}$

squared both sides

$50=(y-5)^{2}+(x-4)^{2}$  -----> equation D

we have the system

$128=(y-8)^{2}+(x-1)^{2}$  -----> equation C

$50=(y-5)^{2}+(x-4)^{2}$  -----> equation D

Solve the system by graphing

The intersection point both graphs is the solution of the system

The solution is the point (9,0)

therefore

The point E is located at (9,0)

see the attached figure to better understand the problem