Parametric Equation

i) Was good, I got t = (1-x)/x
ii) Was good
iii) I found the y-coordinate as y = 1/(1-t)

question iv) The point Q is the intersection of the line x = 1/2 and the straight line joining the origin to the point T. Point R has co-ordinates (1/2, 0). Show that RQ = QP.

I am completely stuck on how to do this question.

Post a better quality image on https://imgur.com/ then post the link here.

The current image is impossible to read.

Yeah sorry for that should be better now.

This question was asked before: https://www.thestudentroom.co.uk/showthread.php?p=75995584

This question was asked before: https://www.thestudentroom.co.uk/showthread.php?p=75995584

Thus, the gradient OT is given by , - I thought from ii) that the gradient m is found by dy/dx = -2t/(1-t)^2 ?

Yeah that's the gradient along the curve.

OT is not along the curve.

I've still been struggling on the question, I used , and hence the line is
When x = 1/2 I plugged that into the parametric equation to find what t was at the point, however t cant be +/- 1. So I dont really know what to do tbh.

When $x=1/2$, what is $y$ in terms of $t$?

Then you know the coordinates of the point Q in terms of $t$. You also know the coords of R and P. So show that the distances RQ = QP

So i just sub x =1/2 into that equation for the line?