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★Given:-
Three points A(5,3), B(3,-2), C(2,-1) .
P is a point such that the area of the quadrilateral PABC is 10 square units.
★To find:-
The locus of P.
★Solution:-
Given that PABC is a quadrilateral.So,On drawing a diagonal AC, ΔABC and ΔPAC are formed.
We know:-
Here,
(x1,y1=5,3)
(x2,y2=3,-2)
(x3,y3=2,-1)
Substituting these values in the above formula,
Therefore,
The area of ΔABC = 7/2 sq.units
Area of ΔPAC :-
⇒ Area of the quadrilateral - Area of the ΔABC
⇒ 10 - 7/2
⇒ 20-7/2
⇒ 13/2
Let (x,y) be the locus of point P.
Now,We have:-
Area of ΔPAC = 13/2
: ⇒ 4x - 3y -11 = ± 13
⇒ 4x - 3y -11 = + 13 and 4x - 3y -11 = - 13
Hence, The locus of P is,
16x² + 9y² - 24xy - 88x + 66y + 290 = 0.
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