M06-09

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.

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|

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

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

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.

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

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 \(\frac{LN * height}{2}\) = \(\frac{2\sqrt{2}*3\sqrt{2}}{2}\)

Area LMN = 6

I think this is a high-quality question and I agree with explanation.

Hi Bunuel,

Why can't the height be 4 and the base be 4 as well? Height being the base of the square to point N, and the base being point M to the left side of the square? So b*h/2 or 4*4/2 = 8

Any side of a triangle can be considered as base and similarly any perpendicular from the opposite vertex to the corresponding base can be considered as height. So, for example, if you consider ML to be base, then the height would be perpendicular from N to ML. None of the sides equals 4 in the triangle MNL and none of the heights is equal to 4. That's why your way is not correct.

Another approach is to use Pick's formula:

Area= M/2 + N - 1,

where
M of lattice points on the boundary placed on the polygon's perimeter
N - lattice points in the interior located in the polygon

M - green dots;
N - blue dots

Area = 6/3 +4 - 1 = 3+4-1 = 6

Find the equation of line passing through (1,3) and (5,1).
(1-3)/(1-5) = (y-3)/(x-1)
or x-2y+1 =0

Find the perpendicular distance of third vertex i.e. (3,5) from this line.

|3-10+1|/sqrt( 2^2+1^2) = 6/(sqrt 5)

Find the distance between (1,3) and (5,1) = 2*sqrt5
Area = 1/2* (6/ sqrt5)* 2 * sqrt5
Area = 6 sq units.

Can you please explain how you found the areas of the 3 smaller triangles?

You can find them using 1/2*base*height. For example, left bottom little triangle has the base of 4 and the height of 2. The area = 1/2*4*2 = 4.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

I think this is a high-quality question and I agree with explanation.