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Maths AS help!

I'm really stuck with this question and keep getting it wrong! Could somebody take me through how you would attempt it step by step?

1. A triangle has vertices (5,0), (q,0) and (10,-1) where q > 5.
Given that the area of the triangle is 4, find the value of q.
Have you drawn a diagram? What have you tried?
what topic is this?
Reply 3
Avatar for MartyO
MartyO
7
Recall the equation for the area of a triangle 12BH\frac{1}{2}BH.

Points (5,0)(5,0) and (q,0)(q,0) lie one the same line. You could pose B=q−5B=q-5, and given that q>5q>5, there is no need for an absolute value.

Points (5,0)(5,0) and (q,0)(q,0) being on the same line (the horizontal axis), HH has to be vertical in order to be perpendicular to BB. H=∣Δy∣=∣−1−0∣=1H=|\Delta y|=|-1-0|=1.

Now that we know the equations for BB and HH, we can solve for qq:

12BH=4\frac{1}{2}BH=4

Spoiler

Reply 4
Avatar for xleo.norx
xleo.norx
OP
13
Original post by samantham999
what topic is this?


Core 1
Reply 5
Avatar for xleo.norx
xleo.norx
OP
13
Original post by Kvothe the Arcane
Have you drawn a diagram? What have you tried?


Yes I did, but it turns out I was over complicating the question haha. I've got the answer now

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