Question

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

Answer 1

★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:

$\leadsto \underline{\boxed{\sf Area\ of\ \triangle = \frac{1}{2} \left| \begin{array}{c c c} \sf {x_1 & y_1 & 1} \\ \sf x_2 & y_2 & 1 \\ \sf x_3 & y_3 & 1 \end{array} \right| }}$

Here,

(x1,y1=5,3)

(x2,y2=3,-2)

(x3,y3=2,-1)

Substituting these values in the above formula,

$:\implies \sf {Area = \left| \begin{array}{c c c} 5 & 3 & 1 \\ 3 & -2 & 1 \\ 2 & -1 & 1 \end{array} \right| }\\\\\\:\implies \sf \dfrac{1}{2} |(5 [(-2)(1) - (1)(-1)] - 3 [(3)(1) - (1)(2)] + (-1) [(3)(-1) - (-2)(2)] ) |\\\\\\:\implies \sf \dfrac{1}{2} | (5(-2+1) - 3(3-2) -1(-3+4)) |\\\\\\:\implies \sf \dfrac{1}{2} | ( -5 - 3 +1 ) |\\\\:\implies \sf \dfrac{1}{2} | -7 |\\\\:\implies \underline{\bold{Area = \frac{7}{2} }}$

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

$:\implies \sf \dfrac{1}{2} \left| \begin{array}{c c c} \sf {x & y & 1} \\ \sf 5 & 3 & 1 \\ \sf 2 & -1 & 1 \end{array} \right| }} = \dfrac{13}{2} \\\\:\implies \sf \dfrac{1}{2} | ( x [(3)(1) - (1)(-1)] - y [(5)(1) - (1)(2)] + (1) [(5)(-1) - (3)(2)] ) | = \dfrac{13}{2} \\\\:\implies \sf \dfrac{1}{2} | x(3+1) - y(5-2) + 1(-5-6) | = \dfrac{13}{2} \\\\:\implies \sf \dfrac{1}{\cancel{2}} | 4x - 3y -11 | = \dfrac{13}{\cancel{2}} \\\\:\implies \sf | 4x - 3y -11 | = 13$

: ⇒ 4x - 3y -11 = ± 13

⇒ 4x - 3y -11 = + 13 and 4x - 3y -11 = - 13

$:\implies \sf ( 4x - 3y -11 )^2 = (+13)(-13)\\\\:\implies \sf 16x^2 + 9y^2+ 2(4x)(-3y) + 2(4x)(-11) + 2(-3y)(-11) + (-11)^2 = -169\\\\ :\implies \sf 16x^2 + 9y^2 - 24xy - 88x + 66y + 121 + 169 = 0\\\\ :\implies \underline{\boxed{\sf 16x^2+ 9y^2- 24xy - 88x + 66y + 290 = 0}}$

Hence, The locus of P is,

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

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Answer 2

★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:

$\leadsto \underline{\boxed{\sf Area\ of\ \triangle = \frac{1}{2} \left| \begin{array}{c c c} \sf {x_1 & y_1 & 1} \\ \sf x_2 & y_2 & 1 \\ \sf x_3 & y_3 & 1 \end{array} \right| }}$

Here,

(x1,y1=5,3)

(x2,y2=3,-2)

(x3,y3=2,-1)

Substituting these values in the above formula,

$:\implies \sf {Area = \left| \begin{array}{c c c} 5 & 3 & 1 \\ 3 & -2 & 1 \\ 2 & -1 & 1 \end{array} \right| }\\\\\\:\implies \sf \dfrac{1}{2} |(5 [(-2)(1) - (1)(-1)] - 3 [(3)(1) - (1)(2)] + (-1) [(3)(-1) - (-2)(2)] ) |\\\\\\:\implies \sf \dfrac{1}{2} | (5(-2+1) - 3(3-2) -1(-3+4)) |\\\\\\:\implies \sf \dfrac{1}{2} | ( -5 - 3 +1 ) |\\\\:\implies \sf \dfrac{1}{2} | -7 |\\\\:\implies \underline{\bold{Area = \frac{7}{2} }}$

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

$:\implies \sf \dfrac{1}{2} \left| \begin{array}{c c c} \sf {x & y & 1} \\ \sf 5 & 3 & 1 \\ \sf 2 & -1 & 1 \end{array} \right| }} = \dfrac{13}{2} \\\\:\implies \sf \dfrac{1}{2} | ( x [(3)(1) - (1)(-1)] - y [(5)(1) - (1)(2)] + (1) [(5)(-1) - (3)(2)] ) | = \dfrac{13}{2} \\\\:\implies \sf \dfrac{1}{2} | x(3+1) - y(5-2) + 1(-5-6) | = \dfrac{13}{2} \\\\:\implies \sf \dfrac{1}{\cancel{2}} | 4x - 3y -11 | = \dfrac{13}{\cancel{2}} \\\\:\implies \sf | 4x - 3y -11 | = 13$

: ⇒ 4x - 3y -11 = ± 13

⇒ 4x - 3y -11 = + 13 and 4x - 3y -11 = - 13

$:\implies \sf ( 4x - 3y -11 )^2 = (+13)(-13)\\\\:\implies \sf 16x^2 + 9y^2+ 2(4x)(-3y) + 2(4x)(-11) + 2(-3y)(-11) + (-11)^2 = -169\\\\ :\implies \sf 16x^2 + 9y^2 - 24xy - 88x + 66y + 121 + 169 = 0\\\\ :\implies \underline{\boxed{\sf 16x^2+ 9y^2- 24xy - 88x + 66y + 290 = 0}}$

Hence, The locus of P is,

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

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