Question

A (5,3) B (3,9) C(2,-1) are 3 points.
If P Is a point such that area of the
quadrilateral PABC is 10 sq units, then the Locus of P Is?

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Answer 1

Step-by-step explanation:

Area of the ΔABC will be :

\frac{1}{2}

2

1

| 5 3 -1 |

| 3 -2 1 |

| 2 -1 1 |

= \frac{1}{2}

2

1

| ( 5 [(-2)(1) - (1)(-1)] - 3 [(3)(1) - (1)(2)] + (-1) [(3)(-1) - (-2)(2)] ) |

= \frac{1}{2}

2

1

| (5(-2+1) - 3(3-2) -1(-3+4)) |

= \frac{1}{2}

2

1

| ( -5 - 3 +1 ) |

= \frac{1}{2}

2

1

| -7 |

= \frac{7}{2}

2

7

Therefore, area of the ΔABC is \frac{7}{2}

2

7

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}

2

13

sq. units

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

2

13

sq. units.

| x y 1 |

\frac{1}{2}

2

1

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

2

13

| 2 -1 1 |

⇒ \frac{1}{2}

2

1

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

2

13

⇒ \frac{1}{2}

2

1

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

2

13

⇒ \frac{1}{2}

2

1

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

2

13

Cancelling \frac{1}{2}

2

1

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