Using distance between two points formula, we can calculate AB = c , BC = a , CA = b . a, b, c represent the lengths of the sides of the triangle ABC.

**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.

With any three non – collinear points *A*(*x*_{1}*
*,* y*_{1})* *, *B *(*x*_{2}* *,*
y*_{2}* *)* *and* C *(*y*_{3}* *,*
y*_{3}* *)* *on a plane, we can form a triangle* ABC*.

Using distance between two points formula, we can calculate *AB *=*
c *, *BC *=* a *, *CA *=* b *.* a*,* b*,* c *represent
the lengths of the sides of the triangle* ABC.*

Using 2*s* = *a* +*b* +*c* , we can calculate
the area of triangle *ABC* by using the Heron’s formula . But
this procedure of finding length of sides of* *Δ*ABC *and then
calculating its area will be a tedious procedure.

There is an elegant way of finding area of a triangle using the
coordinates of its vertices. We shall discuss such a method below.

Let *ABC *be any triangle whose vertices are at *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* to the *x*-axis, respectively.

Clearly *ABQP*, *APRC* and *BQRC* are all
trapeziums.

Now from Fig.5.7, it is clear that

*Area of ΔABC*

*= Area of trapezium ABQP + Area of trapezium APRC − Area of
trapezium BQRC. *

You also know that, *the area of trapezium *

*= 1/2 ×(sum of parallel sides) ×(perpendicular distance between
the parallel sides)*

Therefore, Area of Δ*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 Δ*ABC* is the absolute value
of the expression

= 1/2 {*x* _{1} (*y* _{2} − *y* _{3}
) + *x* _{2}(*y* _{3} − *y*_{1} ) + *x*
_{3} (*y* _{1} − *y*_{2} )} sq.units.

The vertices *A*(*x* _{1} , *y*_{1}
) , *B* (*x*_{2} , *y*_{2} ) and *C* (*x*_{3}
, *y*_{3} ) of Δ*ABC* are said to be “taken in order” if *A*,
*B*, *C* are taken in counter-clock wise direction. If we do this,
then area of Δ*ABC* will never be negative.

**Another form**

The following pictorial representation helps us to write the above
formula very easily.

Area of Δ*ABC* = 1/2 {(*x* _{1} *y*_{2}
+ *x* _{2} *y*_{3} + *x* _{3} *y*_{1}
) −(*x* _{2} *y*_{1} + *x* _{3} *y*_{2}
+ *x* _{1}*y*_{3} )} sq.units.

**Note**

“As the area of a triangle can** **never be negative, we must
take the absolute value, in case area happens to be negative”.

**Progress Check**

The vertices of Δ*PQR* are *P*(0, - 4) , *Q*(3,
1) and *R*(-8, 1)

·
Draw Δ*PQR* on a graph paper.

·
Check if Δ*PQR* is equilateral.

·
Find the area of Δ*PQR.*

·
Find the coordinates of *M*, the mid-point of *QP.*

·
Find the coordinates of *N*, the mid-point of *QR.*

·
Find the area of Δ*MPN.*

·
What is the ratio between the areas of Δ*MPN* and Δ*PQR*
?

**Collinearity of three
points**

If three distinct points *A*(*x* _{1} , *y*_{1}
) , *B*(*x *_{2}* *,* y*_{2}* *)* *and*
C*(*x *_{3}* *,* y*_{3}* *)* *are
collinear, then we* *cannot form a triangle, because for such a triangle
there will be no altitude (height). Therefore, three points *A*(*x* _{1}
, *y*_{1} ) , *B*(*x* _{2} , *y*_{2}
) and *C*(*x* _{3} , *y*_{3} ) will be collinear if the area of Δ*ABC *=0.

Similarly, if the area of ΔABC is zero, then the three points lie
on the same straight line. Thus, three
distinct points *A*(*x*_{1},*y*_{1}), *B*(*x*_{2},*y*_{2}) and *C*(*x*_{3},*y*_{3}) will be collinear if and only if area of ∆ABC = 0.

Tags : Coordinate Geometry Coordinate Geometry

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