derive area of triangle in coordinate geometry
Answers
Answer:
Area of a triangle can be found using three different methods. The three different methods are discussed below
Method 1
When the base and altitude of the triangle are given.
Area of the triangle, A = bh/2 square units
where b and h are base and altitude of the triangle respectively.
Method 2
When the length of three sides of the triangle are given, the area of a triangle can be found using the Heron’s formula.
Therefore, the area of the triangle is calculated using the equation,
A = s(s − a) (s − b) (s − c)−−−−−−−−−−−−−−−−−−−−−√
Where a,b, c are the side lengths of the triangle and s is half of the perimeter
The value of s is found using the formula
s = a + b + c2
Method 3
If the vertices of a triangle are given, first we have to find the length of three sides of a triangle. The length can be found using the distance formula
The procedure to find the area of a triangle when the vertices are known in the coordinate plane is given below.
Let us assume a triangle PQR, whose coordinates P, Q, and R are given as (x1, y1), (x2, y2), (x3, y3) respectively.
Area Of A Triangle in Coordinate Geometry
From the figure, the area of a triangle PQR, lines such as QA←→ , PB←→ and RC←→ are drawn from Q,P and R respectively perpendicular to x – axis.
Now, three different trapeziums are formed such as PQAB, PBCR and QACR in the coordinate plane.
Now, calculate the area of all the trapeziums.
Therefore, the area of ∆PQR is calculated as, Area of ∆PQR=[Area of trapezium PQAB + Area of trapezium PBCR] -[Area of trapezium QACR] —-(1)
Finding Area of a Trapezium PQAB
We know that, the formula to find the area of a trapezium is
Since, Area of a trapezium = (1/2) (sum of the parallel sides)×(distance between them)
Area of trapezium PQAB = (1/2)(QA+PB) × AB
QA = y2
PB =y1
AB = OB – OA = x1-x2
Area of trapezium PQAB = (1/2)(y1+y2)(x1-x2 ) —-(2)
Finding Area of a Trapezium PBCR
Area of trapezium PBCR =(1/2) (PB + CR) × BC
PB =y1
CR = y3
BC = OC – OB =x3-x1
Area of trapezium PBCR =(1/2) (y1+ y3 )(x3-x1) —-(3)
Finding Area of a Trapezium QACR
Area of trapezium QACR = (1/2) (QA + CR) × AC
QA=y2
CR = y3
AC = OC – OA = x3-x2
Area of trapezium QACR =(1/2)(y2+ y3 ) (x3-x2 )—-(4)
Substituting (2), (3) and (4) in (1) gives,
Area of ∆PQR = (1/2)[(y1+y2)(x1-x2 )+(y1+ y3 )(x3-x1)-(y2+ y3 ) (x3-x2 )]
A = (1/2) [x1 (y2- y3 )+x2 (y3-y1 )+x3(y1-y2)]
Special Case:
If one of the vertices of the triangle is origin, then
Area of triangle with vertices are (0,0) P(a,b), and Q(c,d) is
A = (1/2)[0(b – d) + a(d – 0) + c(0 – b)]
A = (ad – bc)/2
If area of triangle with vertices P(x1, y1), Q(x2, y2) and R(x3, y3) is zero, then (1/2) [x1 (y2- y3 )+x2 (y3-y1 )+x3(y1-y2)] = 0 and the points P(x1, y1), Q(x2, y2) and R(x3, y3)are collinear.
Area of a Triangle in Coordinate Geometry Example
Example: What is the area of the ∆ABC whose vertices are A(1,2), B(4,2) and C(3,5)?
Solution:
Using the formula,
A = (1/2) [x1 (y2- y3 )+x2 (y3-y1 )+x3(y1-y2)]
A = (1/2) [1(2 – 5) + 4(5 – 2) + 3(2 – 2)]
A = (1/2) [-3 +12]= 9/2 square units.
Therefore, the area of a triangle ABC is 9/2 square units.