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Area of a Triangle

In your earlier classes, you have studied how to calculate the area of a triangle when its base and corresponding height (altitude) are given. You have used the formula.

Area of a triangle =

 


Let ABC be any triangle whose vertices are A(x1, y1), B(x2, y2) and C(x3, y3). Draw AP, BQ and CR perpendiculars from A, B and C, respectively to the x-axis. Clearly ABQP, APRC and BQRC are all trapezium.

Now from figure, it is clear that

Area of Δ ABC = area of trapezium ABQP + area of trapezium APRC - area of trapezium BQRC.

You also know that the area of a trapezium = (sum of parallel sides) (distance between them)

Therefore,

Area of Δ ABC = (BQ + AP) QP + (AP + CR) PR - (BQ + CR) QR

                    = (y2 + y1) (x1− x2) + (y1 + y3) (x3− x1) − (y2 + y3) (x3 – x2)

                    = [x1(y2− y3) + x2(y3− y1)+ x3 (y1 – y2)]

Thus, the area of Δ ABC is the numerical value of the expression

                    = [x1(y2− y3) + x2(y3− y1)+ x3 (y1 – y2)]


Condition for collinearity of three points :

Let the given points be A(x1,y1) , B(x2, y2) , C(x3 , y3) . If A, B and C are collinear, then,

Area of a Δ ABC = 0 . (i.e.) [x1(y2 − y3) + x2(y3 − y1)+ x3 (y1 − y2)] = 0

And hence , x1(y2 − y3) + x2(y3 − y1)+ x3 (y1 − y2) = 0

 

Example

Find the value of p for which the points (−5,1) , (1,p) and (4, −2) are collinear.

Solution

The given points are A(−5,1),B(1,p) and C(4, −2)
The given points A, B and C are collinear
⇔ x1(y2 − y3) + x2(y3 − y1)+ x3 (y1 − y2) = 0
⇔ (−5).(p + 2) + 1.(−2−1) +4.(1−p) = 0
⇔ (−5p − 10 − 3 + 4 − 4p) = 0
⇔ (−9p − 9) = 0
⇔ 9p =−9 p= −1
Hence p = −1.

 
 




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