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# A (5,3), B (3,-2) are two fixed points, find the equation to the locus of a point P which moves so that the area of the triangle PAB is 9 units.

509 Points
10 years ago

let p be any point (X,Y) then

AB(vector) = -2i - 5j

AP(vector) = (X-5)i + (Y-3)j

area of triangle = (AB*AP)/2                                    (* is cross product)

+or-9  = (-2(Y-3) + 5(X-5)) /2

solve this and get ur ans

Fawz Naim
37 Points
10 years ago

Let the point of which we have to find the locus is C(a,b). Now the other two points are A(5,3),B(3,-2).The area of the triangle formed by these points is given by the formula

1/2 modulus(x1(y2-y3)+x2(y3-y1)+x3(y1-y2)). This area is equal to 9 sq.units

We know the values of x1,x2,x3,y1,y2,y3. Put the values and make an equation in terms of a and b. At the end put x and y at the place of a and b. This is the equation of the locus of point C.