Area of Triangle on coordinate plane

[MoooShuuu]

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How would you go about finding the area and length of sides of a triangle with points A (1,1), B (7,6) and C (3,-6)?

Thanks 👍

 

[UCfan]

Sides can be found by the side formula
AB=(((7-1)^2)+((6-1)^2))=sqrt(36+25)=sqrt(61)
BC=(((7-3)^2)+(6+6)^2))=sqrt(16+144)=sqrt(160)
AC=(((3-1)^2)+(-6-1)^2))=sqrt(4+49)=sqrt(53)
s=(AB+BC+AC)/2
AREA= Sqrt(s*(s-AB)*(s-BC)*(S-AC))
Given points give a messyy calculation!

 

[Streetwolf]

How would you go about finding the area and length of sides of a triangle with points A (1,1), B (7,6) and C (3,-6)?

Thanks 👍

Relocate any point to (0,0) and then relocate the other two accordingly. I choose point A since it's (1,1). If that becomes (0,0) then everything else drops by 1 unit. B becomes (6,5) and C becomes (2,-7). They are all the same distance away from each other as they originally were!! The triangle was just shifted down and left by 1 unit.

Now you can compute the area using the determinant. The determinant figures out the area of the parallelogram of the vectors [6,5] and [2,-7] so you'll want to divide your answer by 2 to get the triangle.

Using det = (a)(d) - (b)(c) and grouping (6 5) above (2 -7) such that a = 6, b = 5, c = 2, and d = -7, you get (6)(-7) - (5)(2) = -52. Area is positive so area = 52. Divide by 2 and get 26, your answer.

Google determinant and look for pictures. You'll see what I mean about parallelograms.

 

[klutzy1987]

Relocate any point to (0,0) and then relocate the other two accordingly. I choose point A since it's (1,1). If that becomes (0,0) then everything else drops by 1 unit. B becomes (6,5) and C becomes (2,-7). They are all the same distance away from each other as they originally were!! The triangle was just shifted down and left by 1 unit.

Now you can compute the area using the determinant. The determinant figures out the area of the parallelogram of the vectors [6,5] and [2,-7] so you'll want to divide your answer by 2 to get the triangle.

Using det = (a)(d) - (b)(c) and grouping (6 5) above (2 -7) such that a = 6, b = 5, c = 2, and d = -7, you get (6)(-7) - (5)(2) = -52. Area is positive so area = 52. Divide by 2 and get 26, your answer.

Google determinant and look for pictures. You'll see what I mean about parallelograms.

Personally i think it is easier to make a square aroud the triangle and then just subract the "extra" right traingles that are formed. What remains is the area of the unknown triangle.

 

[Major Groove]

Personally i think it is easier to make a square aroud the triangle and then just subract the "extra" right traingles that are formed. What remains is the area of the unknown triangle.

Right.

Simpler is better on the DAT for sure.

Another strategy is estimation. Every triangle is half of a rectangle. 1/2 bh.

You don't have to be precise. Ballpark works most of the time. If not, guess and move on.

If I were you I'd practice speed. Learn when to guess, mark it and hopefully there will be time to think later. Just finish the test!! That's your key to success.

 

[MoooShuuu]

Thanks you guys!

 

[Streetwolf]

Personally i think it is easier to make a square aroud the triangle and then just subract the "extra" right traingles that are formed. What remains is the area of the unknown triangle.

No way man, the determinant is so much less work. Takes 20 seconds. Just gotta know how to do it.

 

[klutzy1987]

No way man, the determinant is so much less work. Takes 20 seconds. Just gotta know how to do it.

I guess it depends on the person. For me if the numbers are easy and small I dont mine calculating 3 areas and subtracting. It takes me like 30 secs. Probably a drop slower than usin the determinant but i think def more simple. Only time I use determinant was in a class where the other method wasnt allowed.

 

[MoooShuuu]

Thanks Streetwolf...you explanation really helped me. Very simple!