Find the area of a triangle with vertices P (2,3) Q(4,5) and R(8,9).
Answers
Answer:
Area of a triangle with vertices (x₁, y₁);
(X2, y2) and (x3, Y3) İS
|X1 (Y2 − Y3) + X2 (Y3 − Y1) + X3 (Y1 — Y2) -
2
Hence, substituting the points (X₁, Y₁) =
(5,2);
(X2, Y2) (-9, -3) and (x3, Y3) = =
(-3,-5) in
the area formula, we get area
|5(−3+5) + (−9) (−5 − 2) + (−3)(2+3) -
2
10+63 - 15 2 58 2
29squnits
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Answer:
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Step-by-step explanation:
Using integration finding area of triangle P(2,5)Q(4,7) and R(6,2)
Equation of line PQ
(y−5)=
4−2
7−5
(x−2)⇒y=x+3
Equation of line QR
(y−7)=
6−4
2−7
(x−4)⇒y=
2
−5
x+17
Equation of line PR
(y−5)=
6−2
2−5
(x−2)⇒y=
4
−3
x+
2
13
Area of △PQR
=(area under PQ)+(area under QR)−(area under PR)
=∫
2
4
(x+3)dx+∫
4
6
(−
2
5x
+17)dx−∫
2
6
(
4
−3
x+
2
13
)dx
=[
2
x
2
+3x]
2
4
+[
4
−5x
2
+17x]
4
6
−[
8
3x
2
+
2
13x
]
2
6
=6+3(4−2)+
4
−5
(36−16)+17(6−4)−
8
3
×32−
2
13
×4
=6+6−
4
5
×20+34−12−26
=12−25+34−14=7 sq.unit