From the figure
We have the folloing points
[tex]A(0,0),B(a,c),C(b,-),D(-,-)_{}[/tex]
We need to find the missing points
Since the trapezoid is isosceles the side AB = Side CD
Considering coordinates B and C
let the missing point in C be z then
[tex]\begin{gathered} B=(a,c) \\ C=(b,z) \end{gathered}[/tex]
since Points B and C lie on the same y axis then
z = c.
Hence the coordinate C becomes
[tex]C=(b,c)[/tex]
Next we need to find the missing points in D
let the missing x coordinate be x
let the missing y cordinate be y
Therefore
[tex]D=(x,y)[/tex]
From the Graph,
The point Dlie on the x axis hence the coordinate of y is 0
hence y = 0
Therefore point D will become
[tex]D=(x,0)[/tex]
Finally we are to find the value of x
Recall that the trapezoid is isosceles
hence
[tex]\text{length AB = length CD}[/tex]
Applying the formula for distance between two points
The formula is give as
[tex]\text{Distance =}\sqrt[]{(x_2-x_1)^2-(y_2-y_1)^2}[/tex]
For points AB
A=(0,0), B = ( a,c)
Hence
[tex]\begin{gathered} x_1=0,x_2=a \\ y_1=0,y_2=c \end{gathered}[/tex]
Therefore distance AB is
[tex]\begin{gathered} AB=\sqrt[]{(a-0)^2+(c-0)^2} \\ AB=\sqrt[]{a^2+c^2} \end{gathered}[/tex]
For points CD
C = (b,c), D = (x,0)
Hence,
[tex]\begin{gathered} x_1=b,x_2=x \\ y_1=c,y_2=0 \end{gathered}[/tex]
Therefore, distance CD is
[tex]\begin{gathered} CD=\sqrt[]{(x-b)^2+(0-c)^2} \\ CD=\sqrt[]{(x-b)^2+c^2} \end{gathered}[/tex]
Recall distance AB = distance CD
Hence
[tex]\begin{gathered} AB=CD \\ \sqrt[]{a^2+c^2}=\sqrt[]{(x-b)^2+c^2} \end{gathered}[/tex]
Simplifying further
we will get
[tex]\begin{gathered} a^2+c^2=(x-b)^2+c^2 \\ a^2+c^2-c^2=(x-b)^2 \\ a^2=(x-b)^2 \\ \text{taking square root of both sides} \\ a=x-b \\ x=a+b \end{gathered}[/tex]
Therefore,
The coordinate of the point D is
[tex]D=(a+b,0)[/tex]
