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M06-09

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M06-09 [#permalink]

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Official Solution:

What is the area of a triangle with the following vertices \(L(1, 3)\), \(M(5, 1)\), and \(N(3, 5)\)?

A. 3
B. 4
C. 5
D. 6
E. 7


Make a diagram:

Image

Notice that the area of the blue square is \(4^2=16\) and the area of the red triangle is 16 minus the areas of 3 little triangles which are in the corners (\(2*\frac{2}{2}\), \(4*\frac{2}{2}\) and \(4*\frac{2}{2}\)). Therefore, the area of a triangle LMN is \(16-(2+4+4)=6\).


Answer: D
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Re: M06-09 [#permalink]

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New post 24 Sep 2015, 11:05
a very elegant solution!

i used a longer approach: with Pythagorean theorem I found all the sides and that the traingle is an isosceles triangle with LM=MN, then found the height from M to base LN and finally the area
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Re: M06-09 [#permalink]

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The question can be solved easily by using the area of the triangle by this Area of triangle=(1/2)|D|. where D is determinant.
|1 3 1|
|5 1 1|=1(1*1-5*1)-3(5*1-3*1)+1(5*5-3*1)=1(-4)-3(2)+1(25-3)=-4-6+22=12.
|3 5 1|

If you take positive value of determinant and then and half of that value is the area of triangle.

By this method we could solve the question in 35 odd seconds.

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Re: M06-09 [#permalink]

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viralashara wrote:
The question can be solved easily by using the area of the triangle by this Area of triangle=(1/2)|D|. where D is determinant.
|1 3 1|
|5 1 1|=1(1*1-5*1)-3(5*1-3*1)+1(5*5-3*1)=1(-4)-3(2)+1(25-3)=-4-6+22=12.
|3 5 1|

If you take positive value of determinant and then and half of that value is the area of triangle.

By this method we could solve the question in 35 odd seconds.



would u pls explain how did u get 1 in the right collumn?

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Re: M06-09 [#permalink]

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BelalHossain046 wrote:
viralashara wrote:
The question can be solved easily by using the area of the triangle by this Area of triangle=(1/2)|D|. where D is determinant.
|1 3 1|
|5 1 1|=1(1*1-5*1)-3(5*1-3*1)+1(5*5-3*1)=1(-4)-3(2)+1(25-3)=-4-6+22=12.
|3 5 1|

If you take positive value of determinant and then and half of that value is the area of triangle.

By this method we could solve the question in 35 odd seconds.



would u pls explain how did u get 1 in the right collumn?


Standard formula of Co-ordinate geometry.

If there are three points (a,b),(c,d) and (e,f)
The Area of triangle formed by these three points will be given by
1/2 *|a b 1|
|c d 1|
|e f 1|
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New post 23 Oct 2015, 10:19
1. Calculate the equation of line connecting any 2 points (I used (1,3) and (3,5): Eq: x+2y=7)
2. Calculate the perpendicular distance from the 3rd point to the line (6/sq root (5))
3. Use formula for area of the triangle

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Re: M06-09 [#permalink]

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The area of triangle when 3 vertices are given by

= 1/2 [ x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]

x1,y1 = L(1,3 )
x2,y2 = M(5,1)
x3,y3 = N(3,5)

Substituting the 3 values in the formula we get

=> 1/2 [ 1 ( 1 -5 ) + 5( 5 - 3 ) + 3 ( 3 - 1 )]
= 1/2 [ -4 + 10 + 6 ] = 12/2 = 6

the tricky here is to remembering the correct sequence of the formula.
my short cut to remember this formula is to do a circular rotate first three vertices ( x1,y2,y3) by 1 to get second one ( x2,y3,y1) and do the rotate again to get the third one ( x3,y1,y2). OR you can form a 3x3 matrix to remember well.
1/2 [1,2,3 + 2,3,1 + 3,1,2].
Not sure if this is the wise approach but I could solve faster.

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M06-09 [#permalink]

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New post 17 Jun 2016, 09:21
1/2 *modulus of |x1-x2 x1-x3|
|y1-y2 y1-y3|
simple formula for finding area of triangle when 3 vertices (x1,y1);(x2,y2);(x3,y3) are given. | | indicates determinant.

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Re: M06-09 [#permalink]

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New post 23 Jul 2016, 10:16
Can we do it like this ?

height will be 5 units as seen from the figure.

And the base will be LM which can be found by distance formula : 2\sqrt{5}

so area 1/2*5*2\sqrt{5} = 5\sqrt{5} (close to 6)

Is the approach right?
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New post 30 Jul 2016, 10:53
Step 1: Calculate each side with distance formula

LM=2\sqrt{5} MN=2\sqrt{5} and LN= 2\sqrt{2}



Lets quickly use herons formula.

S= (2\sqrt{5}+\sqrt{5}+2\sqrt{2})/2

S=2\sqrt{5} +\sqrt{2}

Area of triangle= \sqrt{S(S-a)(S-b)(S-c)} = \sqrt{36} = 6

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Re: M06-09 [#permalink]

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New post 07 Aug 2016, 18:23
Determinant approach, Brilliant solution

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Re: M06-09 [#permalink]

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New post 28 Oct 2016, 20:12
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Hey everyone,

This is really a new golden formula to me.

"The area of triangle when 3 vertices are given by

= 1/2 [ x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]"

Just have a concern that if x1 or y3 is negative, or y2-y3 negative, then we just apply the negative values into this formula?

Thanks alot

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New post 20 Nov 2016, 04:55
I could solve it very quickly after I realised it is an isosceles triangle with sides sqrt(20), sqrt (20), sqrt (8). I simply used the formula to calculate the area of an isosceles triangle, which is [ a/4 * sqrt (4 * a^2 - c^2) ] such that side a = side b.

maybe it worked here, but i think the determinant approach, the square approach or the area of triangle using vertices approach are the most standard ones for solving such questions.

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Re: M06-09 [#permalink]

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New post 03 May 2017, 06:24
Hi All,

We have to draw first the figure.

Then apply the formula : \(\frac{(Base*Height)}{2}\)
\((LN*height)/2\)

Since LN is the combined length of two diagonals of a unit square, he have that LN = \(\sqrt{2}+\sqrt{2}\)
Since the Height is the combined length of three diagonals of a unit square we get : height : \(\sqrt{2}+\sqrt{2}+\sqrt{2}\)

So \(\frac{LN * height}{2}\) = \(\frac{2\sqrt{2[}{square_root]*3[square_root]2}/2}\)

Area LMN = 6

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Re: M06-09 [#permalink]

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viralashara wrote:
The question can be solved easily by using the area of the triangle by this Area of triangle=(1/2)|D|. where D is determinant.
|1 3 1|
|5 1 1|=1(1*1-5*1)-3(5*1-3*1)+1(5*5-3*1)=1(-4)-3(2)+1(25-3)=-4-6+22=12.
|3 5 1|

If you take positive value of determinant and then and half of that value is the area of triangle.

By this method we could solve the question in 35 odd seconds.

Thank you for reminding this approach. I had totally forgotten about it. I'd like to add 1 more thing for everybody to understand..

Determinant of a matrix can be found by this general method..

|a b c|
|d e f |
|h i j|

= a*(e*j - i*f) - d*(b*j - I*c) + h*(b*f - e*c)

Note: You can take a, d, & h outside or a, b, & c; or b, e, & i; or c, f, & j; and so on i.e either a row or a column and subtract the cross-multiple of the rest as performed above.
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Re: M06-09 [#permalink]

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New post 21 Nov 2017, 00:10
Clever approach.

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Re: M06-09 [#permalink]

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New post 29 Dec 2017, 03:29
dharan wrote:
The area of triangle when 3 vertices are given by

= 1/2 [ x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]

x1,y1 = L(1,3 )
x2,y2 = M(5,1)
x3,y3 = N(3,5)

Substituting the 3 values in the formula we get

=> 1/2 [ 1 ( 1 -5 ) + 5( 5 - 3 ) + 3 ( 3 - 1 )]
= 1/2 [ -4 + 10 + 6 ] = 12/2 = 6

the tricky here is to remembering the correct sequence of the formula.
my short cut to remember this formula is to do a circular rotate first three vertices ( x1,y2,y3) by 1 to get second one ( x2,y3,y1) and do the rotate again to get the third one ( x3,y1,y2). OR you can form a 3x3 matrix to remember well.
1/2 [1,2,3 + 2,3,1 + 3,1,2].
Not sure if this is the wise approach but I could solve faster.



Hi could you clarify this.

In the formuar you have written \(\frac{1}{2}* [ x1 (y2- y3) + x2(y3-y1) + x3(y1-y2)\)

If the terms were interchanged like this \(\frac{1}{2} * [ x1(y2-y1) + x2( y1-y3) + x3(y3-y2) ]\)

Would it matter, what parameters govern this order.
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Re: M06-09 [#permalink]

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New post 31 Dec 2017, 07:08
i solved using the traditional approach by finding the length of all sides. But great soln by Bunuel.

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Re: M06-09   [#permalink] 31 Dec 2017, 07:08
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