One Formula a special case of a more general Formula

An understanding of what you are doing in maths rather than just memorising and slavishly applying formulae in a given format to a particular situation allows one to draw on learning and information from different different parts of your course and apply it to solving problems. For example on page 18 of the F&T booklet dealing with co-ordinate geometry there is a formula for the area of a triangle which has one vertex at the origin (0,0) and the others at ( x1,y1) and (x2,y2)

Area of triangle with vertex at (0,0) = 1/2 |x1y2-x2y1|

This can be used to find the area of any triangle for which we know the co-ordinates of all three vertices even if none of them are at the origin by simply finding the translation that would bring one of them to (0,0) and finding the corresponding coordinates of the other two under the same translation.

The above formula is actually a special case of a more general formula for the area of a triangle, which is not on the leaving certificate course.

Given the coordinates of the three vertices of a triangle ABC, the area is given by

 Area of triangle with known coordinates of vertices

where Ax and Ay are the x and y coordinates of the point A etc..

If one vertex is at the origin point at (0,0), the above formula reduces to the special case given on p18 of F&T booklet.

Formula for area of triangle

This formula allows you to calculate the area of a triangle when you know the coordinates of all three vertices. It does not matter which points are labelled A,B or C, and it will work with any triangle, including those where some or all coordinates are negative.

Looking at the formula above, you will see it is enclosed by two vertical bars like this: Area = |A|

The two vertical bars mean “absolute value”. This means that it is always positive even if the formula produced a negative result. Polygons can never have a negative area.

To use the previous formula for a triangle with a vertex at (0,0) we could translate say the vertex at A to (0,0). To do this the x and y coordinates are each reduced by 15. Doing the same for points B and C gives translated coordinates at (8,15) and (35,10) respectively. Substituting these into the formula Area = 1/2 |x1y2-x2y1| gives 1/2|(8)(10)-(35)(15)| = 222.5 as expected.

If the area comes out to be zero, it means the three points are collinear. They lie in a straight line and do not form a triangle.

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