MEI Core 1 Coordinate Geometry

If anyone can help me with the answers to these, with solutions?!

The coordinates of four points are P(-2,-1) Q(6,3) R(9,2) and S(1,-2)
i) Calculate the gradients of the lines PQ, QR, RS, SP
ii) What is the name given to the quadrilateral PQRS
iii) Calculate the length SR
iv) Show that the equation of SR is 2y=x-5 and find the equation of the line L through Q perpendicular to SR
v) Calculate the coordinates of the point T where the line L meets SR
vi) Calculate the area of the quadrilateral PQRS

How far did you get? Could you share your working out?

just use the formulas...

Well for part i, I got the answers
PQ and RS - Gradient 0.5
QR and SP - Gradient 1/-3

ii) It's a parallelogram

iii) I got SR = root 80

iv) y = -2x -9

But this i'm not sure on which is why I don't think I can get the next part

Almost, well done

The -9 in your equation for L should be a +9. Can you see why/how?

I can indeed because I substituted +6 not -6?
Thank you, I shall try and work out the rest now and get back to you

Mmm. that may be it but I'm not sure if you've done it for the y values a well then.

Remember that if you have one point (a,b) and the gradient,m, then the eqn is (y-b)=m(x-a).

I used this formula and that is how I got y=-2x-9

gradient = -2/1, through point (6,3)

y-3= -2/1x-12/1

times 1

y-3 = -2x-12

+3

y = -2x -9

Sorry, I have misled you. It is meant to be y=-2x + 15.

The step where you got -2x - 12 should actually be +12. Because on that side, it's m(x-b), which is -2(x-6).

Again, the formula is (y-b)=m(x-a) when you have the gradient m and the point (a,b).