# 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(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}). 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 Î” 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 Î” ABC = (BQ + AP) QP + (AP + CR) PR - (BQ + CR) QR

= (y_{2} + y_{1}) (x_{1}âˆ’ x_{2}) + (y_{1} + y_{3}) (x_{3}âˆ’ x_{1}) âˆ’ (y_{2} + y_{3}) (x_{3} â€“ x_{2})

= [x_{1}(y_{2}âˆ’ y_{3}) + x_{2}(y_{3}âˆ’ y_{1})+ x_{3} (y_{1} â€“ y_{2})]

Thus, the area of Î” ABC is the numerical value of the expression

= [x_{1}(y_{2}âˆ’ y_{3}) + x_{2}(y_{3}âˆ’ y_{1})+ x_{3} (y_{1} â€“ y_{2})]

**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 Î” ABC = 0 . (i.e.) [x1(y2 âˆ’ y3) + x2(y3 âˆ’ y1)+ x3 (y1 âˆ’ y2)] = 0

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

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

The given points are A(âˆ’5,1),B(1,p) and C(4, âˆ’2)

The given points A, B and C are collinear

â‡” x1(y2 âˆ’ y3) + x2(y3 âˆ’ y1)+ x3 (y1 âˆ’ y2) = 0

â‡” (âˆ’5).(p + 2) + 1.(âˆ’2âˆ’1) +4.(1âˆ’p) = 0

â‡” (âˆ’5p âˆ’ 10 âˆ’ 3 + 4 âˆ’ 4p) = 0

â‡” (âˆ’9p âˆ’ 9) = 0

â‡” 9p =âˆ’9 p= âˆ’1

Hence p = âˆ’1.