Math homework help

[becomingagodo]

Sorry for asking for help, however this problem is weird and the method to solve it is not in the textbook. Look in Heinmann modular mathematics C1 if you don't believe me.

A(-1,-2) B(7,2) C(k,4)
calculate K
Now I know K=6 because I cheated and looked.

How do you do this?

Sketching the graph would easily get me the anwser, however it said to calculate.

[andyandy]

Sorry for asking for help, however this problem is weird and the method to solve it is not in the textbook. Look in Heinmann modular mathematics C1 if you don't believe me.

A(-1,-2) B(7,2) C(k,4)
calculate K
Now I know K=6 because I cheated and looked.

How do you do this?

Sketching the graph would easily get me the anwser, however it said to calculate.

does the question give you any more information? Are the three points supposed to form a right angled triangle (say)? Without any extra information k could be anything surely, as there has been no relationship specified between A, B and C.

[OnlyTellsTruths]

Wouldn't k=11?

[OnlyTellsTruths]

Did I just get tricked into doing someones work... I take back the previous answer of 11, I now agree it is 6.

[maxpower1227]

Yeah, for the points to be collinear k=11. If you say the answer is k=6, I have to wonder what the question was.

[maxpower1227]

Ok, after looking at it again, it appears triangle ABC is to be a right triangle with the right angle at B. To solve this, you must set the dot product of vectors BA and BC equal to zero.

So:

(Bx-Ax)(Bx-Cx) + (By-Ay)(By-Cy) = 0
(7-(-1))(7-k) + (2-(-2))(2-4) = 0
8(7-k) + (4)(-2) = 0
8(7-k) = 8
7-k = 1
k = 6

[TV's Frank]

The question was probably, "If k were equal to 6, what would k equal?"

And they say math education has gone downhill!

[becomingagodo]

Ok, after looking at it again, it appears triangle ABC is to be a right triangle with the right angle at B. To solve this, you must set the dot product of vectors BA and BC equal to zero.

So:

(Bx-Ax)(Bx-Cx) + (By-Ay)(By-Cy) = 0
(7-(-1))(7-k) + (2-(-2))(2-4) = 0
8(7-k) + (4)(-2) = 0
8(7-k) = 8
7-k = 1
k = 6
That saved me some time.

I wonder why they gave a problem you can't solve by reading the textbook. We don't do vectors intill C2.

[DazzaD]

Pythagorean theorem

or

look at the gradients of the lines that are supposed to 'form' the right angle

or some similar way of thinking that only requires GCSE Maths

All of that chapter is really just a revision of GCSE Maths (with the possible exception of being told about the relationship between the gradients of lines that are perpendicular.. not everyone gets taught that by their GCSE teachers).