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 :
`text (Area of a triangle) = 1/2 xx text (base) xx text (altitude )`
Now, if the coordinates of the vertices of a triangle are given. Well, you could find the lengths of the three sides using the distance formula and then use Heron’s formula.
But this could be tedious, particularly if the lengths of the sides are irrational numbers. Let us see if there is an easier way out.
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 trapezia (see Fig. 7.13).
Now, from Fig. 7.13, it is clear that area of
`Delta ABC = text ( area of trapezium ) ABQP + text (area of trapezium) APRC – text (area of trapezium) BQRC`.
You also know that the
`text (area of a trapezium ) = 1/2xx "sum of parallel sides" xx "distance between them"`
Therefore, Area of
`Delta ABC =1/2 (BQ +AP) QP +1/2 (AP +CR) PR -1/2 (BQ +CR) QR`
` =1/2 (y_2 +y_1) (x_1 - x_2) +1/2 (y_1 +y_3) (x_3 -x_1) -1/2 (y_2 +y_3) (x_3 -x_2 )`
` =1/2 [x_1 (y_2 -y_3) +x_2 (y_3 -y_1) + x_3 ( y_1 -y_2) ]`
Thus, the area of `Delta ABC` is the numerical value of the expression
`color{red}{= 1/2 [ x_1 (y_2 - y_3) +x_2 (y_3 -y_1) + x_3 (y_1 - y_2)]}`
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 :
`text (Area of a triangle) = 1/2 xx text (base) xx text (altitude )`
Now, if the coordinates of the vertices of a triangle are given. Well, you could find the lengths of the three sides using the distance formula and then use Heron’s formula.
But this could be tedious, particularly if the lengths of the sides are irrational numbers. Let us see if there is an easier way out.
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 trapezia (see Fig. 7.13).
Now, from Fig. 7.13, it is clear that area of
`Delta ABC = text ( area of trapezium ) ABQP + text (area of trapezium) APRC – text (area of trapezium) BQRC`.
You also know that the
`text (area of a trapezium ) = 1/2xx "sum of parallel sides" xx "distance between them"`
Therefore, Area of
`Delta ABC =1/2 (BQ +AP) QP +1/2 (AP +CR) PR -1/2 (BQ +CR) QR`
` =1/2 (y_2 +y_1) (x_1 - x_2) +1/2 (y_1 +y_3) (x_3 -x_1) -1/2 (y_2 +y_3) (x_3 -x_2 )`
` =1/2 [x_1 (y_2 -y_3) +x_2 (y_3 -y_1) + x_3 ( y_1 -y_2) ]`
Thus, the area of `Delta ABC` is the numerical value of the expression
`color{red}{= 1/2 [ x_1 (y_2 - y_3) +x_2 (y_3 -y_1) + x_3 (y_1 - y_2)]}`