# Quant reasoning problem

#### predentgal

In a 2 dimensional coordinate system, point A=(3, -6), point B= (1,1) and point C = (7 ,6). Find the area of triangle ABC.

The answer is 26

I thought you had to use the distance formua to find the the distance of the base and height but i'm getting some crazy radical numbers when I use the distance formula so I know there must be some other way to answer this. I hate mathhhh!! If anyone knows how to do this please help. One more question too:
Which is the third term in the expansion of (x + 3)^5

I don't even know what the question is asking!!!

The answer is 90x^3

Last edited:

#### UndergradGuy7

10+ Year Member
For the triangle, I think you need to use the distance formula a^2 + b^2 = c^2 and solve for c the distance of each side. Do this for all 3 sides then use Heron's formula for the area of a triangle. http://en.wikipedia.org/wiki/Heron%27s_formula

For the second one, you need to expand the term like if you wrote it out as (x+3)(x+3)... and multiplied everything out they want to know the 3rd term. Nobody multiplies it out, they learn the binomial theorum formula.
http://en.wikipedia.org/wiki/Binomial_theorem

Maybe someone else can show the work or explain it better.

#### Ibracadabra

In a 2 dimensional coordinate system, point A=(3, -6), point B= (1,1) and point C = (7 ,6). Find the area of triangle ABC.

The answer is 26

I thought you had to use the distance formua to find the the distance of the base and height but i'm getting some crazy radical numbers when I use the distance formula so I know there must be some other way to answer this. I hate mathhhh!! If anyone knows how to do this please help. One more question too:
Which is the third term in the expansion of (x + 3)^5

I don't even know what the question is asking!!!

The answer is 90x^3
I doubt you have to know this formula but...
when you have three coordinates (x&#8321;,y&#832 ,(x&#8322;,y&#8322 ,(x&#8323;,y&#8323 , the area of the triangle is 1/2|(x&#8321;y&#8322;+x&#8322;y&#8323;+x&#8323;y¹)-(x&#8322;y&#8321;+x&#8323;y&#8322;+x&#8321;y&#8323 |. (you need the absolute values) I've had these type of problems in high school and unfortunately I forgot how to derive it.. I'm sure someone else can do that for you.
If you've learned vector, I think you can do it more simple.. but for now that equation is what I can tell you.

The second question is asking when you unfold/expand (x + 3)^5, you will get sth like ax^5+bx^4+cx^3+dx^2+ex+f.. it's asking for the c value in this case. I'm quite sure you can do these questions without actually unfolding them.. I'm quite sure you can use probability to solve this question or multiplying out just the ones that make ~x^3..
I'm sorry for not being much of help, but I have to go now so I'll see what other people says about this one! #### herkulease

10+ Year Member
the 2nd question is merely asking you to do simply the following
(x+3)(x+3)(x+3)(x+3)(x+3)

I don't know what the other answer choices are but if its not one positive and 1 negative like (x+y)(x-y). you'll essentially get a count down from the exponent down.

for example (x+1)^10 you'll end up with 11 terms when you expanded. the 1st term would be x^10 then x^9, x^8..... the 11th term is 1^10 or 1.

You can quickly eliminate choices that way. By the way what were the choices. I know my explaination is kinda confusing. I'll try to think it over and will make edits if I could. It something I just know.

for 1st question use this formula.

http://www.mathopenref.com/coordtrianglearea.html

wikipedia has it. but its rather easy to confuse the terms.

#### wantVCUdental

In a 2 dimensional coordinate system, point A=(3, -6), point B= (1,1) and point C = (7 ,6). Find the area of triangle ABC.

The answer is 26
this is probably the more challenging of the math questions, it's challenging b/c of all the calculation req.
1. you need to decide on a line as the base, pick any one (AB, BC, AC)
2. then you find the equation of that line
3. then here comes the tricky part, use the slope of the line you found, take the negative reciprical and then use that as the slope of your height which attaches on to the third point that you have not used so far
4. then find equation of that line (height)
5. then find the intersection of that height with your base
6. now it's just finding the length of your base and height (which is all easy now)
7. now you multiple the distance together, divide by 2

following what i said above, try this:
A=(3, -6), point B= (1,1) and point C = (7 ,6). Find the area of triangle ABC.
pick A, C
the slope = 12/4=3 using this, now we find the equation representing the line (AC) 6=3(7)+c where c=-15
so the equation of AC is y=3x-15 (this is our base)

now taking the negative reciprocal of the slope we found which becomes -1/3.
and using the point we haven't touched yet (B)
1=-1/3 (1)+c where c=4/3, so our height can be represented by
y=-1/3x+4/3 (this is our height)

now where do those intersect?
y=3x-15
y=-1/3x+4/3
let's equate them
3x-15=-1/3x+4/3
x=4.9 y=0.3
*note, the answer in you gave probably kept everything irrational, cuz what follows i tried to keep everything simple and did some rounding in the calculation and i got something very close but not exactly 26

okay, so...
now we find the distance between AC sqrt((12^2)+(4^2)) =12.64 (approx)
distance between base and point B=sqrt((0.7^2)+(3.9^2))=3.96
now we plug in and solve Area=base x height/2
Area=3.96x12.64/2=25
Hope this helps

#### Streetwolf

##### Ultra Senior Member
10+ Year Member
7+ Year Member
I doubt you have to know this formula but...
when you have three coordinates (x&#8321;,y&#832 ,(x&#8322;,y&#8322 ,(x&#8323;,y&#8323 , the area of the triangle is 1/2|(x&#8321;y&#8322;+x&#8322;y&#8323;+x&#8323;y¹)-(x&#8322;y&#8321;+x&#8323;y&#8322;+x&#8321;y&#8323 |. (you need the absolute values) I've had these type of problems in high school and unfortunately I forgot how to derive it.. I'm sure someone else can do that for you.
If you've learned vector, I think you can do it more simple.. but for now that equation is what I can tell you.
I'm not going to check that for accuracy but it deals with the determinant with three different points. An easier way (and I've answered this particular question about 10 times now) is to set one of the coordinates to (0,0) which would eliminate a lot of that work. For instance if (x3, y3) became (0,0) it would simplify your equation to:

1/2 |(x1y2) - (x2y1)|

And the stuff in absolute value brackets is the determinant of two points.

So the easiest point to move to (0,0) is the point (1,1). Do this by moving the triangle left and down by 1.

The other points become (6,5) and (2, -7).

The equation reads 1/2 |(6*-7) - (5*2)| = 1/2 |-52| = 1/2(52) = 26.