Question

.27. A (5,3), B (3,-2), C(2, -1) are three points. If P is a variable point such that the
area of the quadrilateral PABC is 10 sq.units, then the locus of P is ​

Answer 1

Given data :

A (5,3), B (3,-2), C(2, -1) are three points.

P is a variable point such that the  area of the quadrilateral PABC is 10 sq.units

To find :

The locus of point P.

Step by step calculation :

As PABC is a quadrilateral [Look at the attachment given below], On drawing a diagonal AC, ΔABC and ΔPAC are formed.

Let (x,y) be the coordinates of the locus point P.

Area of the ΔABC will be :  

$\frac{1}{2}$ | 5       3       -1 |

  | 3      -2       1 |

  | 2      -1        1 |

= $\frac{1}{2}$ | ( 5 [(-2)(1) - (1)(-1)] - 3 [(3)(1) - (1)(2)] + (-1) [(3)(-1) - (-2)(2)] ) |

= $\frac{1}{2}$ | (5(-2+1) - 3(3-2) -1(-3+4)) |

= $\frac{1}{2}$ | ( -5 - 3 +1 ) |

= $\frac{1}{2}$ | -7 |

= $\frac{7}{2}$

Therefore, area of the ΔABC is $\frac{7}{2}$ sq. units.

Then, the area of the remaining portion, i.e,

Area of the ΔPAC = Area of the quadrilateral - Area of the ΔABC

                              = 10 - (7/2)

                              = $\frac{13}{2}$ sq. units

Now, to find the locus of P(x,y), equate the area of triangle PAC fomat to its area $\frac{13}{2}$ sq. units.

   | x       y        1 |

$\frac{1}{2}$  | 5      3        1 |     =  $\frac{13}{2}$

    | 2      -1        1 |

⇒ $\frac{1}{2}$ | ( x [(3)(1) - (1)(-1)] - y [(5)(1) - (1)(2)] + (1) [(5)(-1) - (3)(2)] ) | = $\frac{13}{2}$

⇒ $\frac{1}{2}$ | x(3+1) - y(5-2) + 1(-5-6)  | = $\frac{13}{2}$

⇒ $\frac{1}{2}$ | 4x - 3y -11 | = $\frac{13}{2}$

Cancelling  $\frac{1}{2}$ on both sides, we get :

⇒ | 4x - 3y -11 | = 13

⇒ 4x - 3y -11 = ± 13

So, the equations are 4x - 3y -11 = + 13 and 4x - 3y -11 = - 13, and the locus will be :

⇒ ( 4x - 3y -11 )² = (+13)(-13)

⇒ 16x² + 9y² + 2(4x)(-3y) + 2(4x)(-11) + 2(-3y)(-11) + (-11)² = -169

⇒ 16x² + 9y² - 24xy - 88x + 66y + 121 + 169 = 0

∴ 16x² + 9y² - 24xy - 88x + 66y + 290 = 0

Therefore, the locus of point P(x,y) is

16x² + 9y² - 24xy - 88x + 66y + 290 = 0

Learn more :

1) find the equation of locus of point p the distance of p from origin is twice the distance of p from A{1,2}​

https://brainly.in/question/10701069

2) Let p be the point (1,0) and Q a point on the locus y² = 8x . the locus of the mid point of PQ is

https://brainly.in/question/7423673

Answer 2

Answer:

16x² + 9y² - 24xy - 88x + 66y + 290 = 0