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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.

```
9 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
```
9 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.
```
9 years ago
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