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Please help me with this rather large C1 Geometry Question. I'm going crazy.

Question:
IMG_20170114_223920.jpg
For part A) I subbed in x=-3 and x=1 into the equation to find the y values, giving me y=-18 and y=18

However, I don't know how on earth I would go on from there for part A to answer the question. Please could you help me and explain what to do and why?

For part B, I guess it would because they have the same gradient, but please could you help me with what to do for that one?

And just explain how to do c, d, e if you could please, but im mainly focusing on parts A and B atm.

Thanks a lot, I appreciate it. This question has got me so stressed.
(edited 7 years ago)
Reply 1
Avatar for fefssdf
fefssdf
20
Actuallt hate C1
Original post by blobbybill
...


Part A - you have 2 points. Construct a line equation going through both of them by taking the gradient between the two points use one of the points for y−y1=m(x−x1)y-y_1=m(x-x_1)

Part B - differentiate the function and show that y′(−3)=y′(1)y'(-3)=y'(1)

Part C - just calculate the distance PR using Pythagoras' Theorem.

Part D - same approach as in C.

Part E - well you can explain the reason using the information given right before part C then find the area by doing 12(PQ)(QR)\frac{1}{2}(PQ)(QR)
Reply 3
Avatar for mani119
mani119
7
a) use (y-y1/x-x1) to find gradient
P(-3,-18) Q(1,18)
so gradient=(-18-18/-3-1)=9
using y-y1=m(x-x1)
y-18=9(x-1)
y=9x+9

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