Give the coordinates for point P in the following right trapezoid without using any new variables

Answer with explanation:
→→Naming of trapezoid =O P QR=O(0,0),P(x,0),Q(a,b) and R (0,c)
Where the Coordinate of vertices of Trapezoid P =(x,0)
→Slope between two points (m,n) and (c,d) is given by
[tex]\frac{n-d}{m-c}[/tex]
→Slope of segment OP,where, O(0,0),and P(x,0),using the above formula
[tex]\frac{0-0}{x-0}=\frac{0}{x}[/tex]
Similarly, Slope of segment OR,where, O(0,0),and R(0,c),using the above formula
[tex]\frac{c-0}{0-0}=\frac{c}{0}[/tex]
→Side OP and OR are Perpendicular to each other.
Slope of segment OP × Slope of segment OR= -1
[tex]\frac{0}{x} \times \frac{c}{0}= -1\\\\ -x=c\\\\ x= -c[/tex]
But this point lies on positive side of x axis, so the coordinate of vertices of point P should be (c,0).
There is another way of solving this problem.
Draw perpendicular from Vertex Q(a,b)on X axis.Coordinates of P will become (a,0).
Now, draw perpendicular from vertex R on side PQ.
Opposite sides of rectangle are equal.
OP=MR=a
OR=MP=c
OR ⊥ OP
Product of their slopes ,that is slope of OR and OP= -1
[tex]\frac{0}{a} \times \frac{c}{0}= -1\\\\ -a=c\\\\ a= -c[/tex]
But point ,P lies on positive side of x axis.
So, coordinates of point P = (c,0)
Option B (c,0)