# choo choo, all aboard the resit train ;_;

##### 2017-05-18 10:29:38 +0000, 1 year and 4 months ago

Ok, I’m pretty sure the maths exam was a total trainwreck. As in, I didn’t even part c -> b of question 9. There were some difficult questions, for example (if I’m remembering this properly);

Point A, B and C lie on the co-ordinates (-2, 8), (5, -6) and (k, k+1), respectively, given that angle BCA is a right angle, find the possible values of k.

Seems easy enough right? Welp, most people who came out of the exam also had no idea how to tackle it either, I’m not gonna’ lie I totally bailed out of answering it when I first laid eyes upon it, so lets see if I can answer it now.

Since we know that AC is perpendicular to BC, then if we find the gradient of each line AC, and BC, we might end up with something workable.

The perpendicular rule states that $m=\frac{-1}{m}$, so $AC \times BC = -1$, so we can multiply them together;

This is looking better, I guess we could multiply the denominator to get $-((k+2...$;

Hey that looks like a nice quadratic to me, lets solve for k.

Nice nice, time to factorise, factors of 44 are 2, 22 and 4, 11. 11 and 4 look good since 2(4)=8, so that looks like -11+8 = 3, so that’s our match.

Thus the values of $k$ are $\frac{11}{2}$ and $-4$.

Wish I’d spent less time dicking about on other questions and spent some time on this question for the 5 marks, rip my grades. Good question though. I’ll probably be resitting the exam next year though.

Return?

### Comments

there's some rate-limiting going on with my comment provider atm, so i'd reccommend copying your message incase it gets blocked and retrying