Question

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NDA 2014 (||)​

Answer 1

Answer:

Coordinates of point P = (7 , 2)

Explanation:

Let the coordinates of P be (x , y)

Given that PA = PB

PA = (x , y) (3 , 4)

PB = (x , y) (5 , -2)

By distance formula,

PA = $\rm\sqrt{(3-x)^2+(4-y)^2}$

PB = $\rm\sqrt{(5-x)^2+(-2-y)^2}$

So,

$\rm\sqrt{(3-x)^2+(4-y)^2} = \rm\sqrt{(5-x)^2+(-2-y)^2}$

Squaring on both sides,

(3 - x)² + (4 - y)² = (5 - x)² + (-2 - y)²

= 9 + ⅹ² - 6ⅹ + 16 + y² - 8y = 25 + ⅹ² - 10ⅹ + 4 + y² + 4y

= -6ⅹ - 8y = -10x + 4 + 4y

= 4ⅹ - 12y = 4 ［divide each term by 4］

= ⅹ - 3y = 1 → eqn 1

Area of ∆PAB = 10

Area of triangle of (3 , 4) (5 , -2) (x , y)

$= \sf \dfrac{1}{2} \bigg[x_1(y_2 - y_3) + x_2(y_3- y_1) + x_3(y_1 - y_2) \bigg] = 10 \\ \\ = \sf \dfrac{1}{2} \bigg[3( - 2 - y) + 5(y- 4) + x(4 + 2) \bigg] = 10 \\ \\ = \sf - 6 + 2y - 20 + 6x = 20 \\ \\ \sf = 2y + 6x = 46 \\ \\ \sf = 3 x + y = 23 \: \: (eqn \: 2)$

On solving eqn 1 and 2 we get,

x = 7

y = 2

∴ Coordinates of point P = (7 , 2)

Answer 2

Answer:

A(3,4)

B(5,-2)

PA=PB

Let coordinates of P(x,y)

Area of triangle PAB = 10 sq.units

Let,

x1 = 3

x2 = 5

x3 = x

y1 = 4

y2 = -2

y3 = y

$area = \frac{1}{2} (x1(y2 - y3) + x2(y3 - y1) + x3(y 1- y2))$

$10 = \frac{1}{2} (3( - 2 - y) + 5(y - 4) + x(4 + 2) \\ 20 = - 6 - 3y + 5y - 20 + 6x \\ 20 = - 26 + 2y + 6x \\ 6x + 2y = 46 \\ 3x + y = 23 \: \: \: \: \: \: \: \: \: \: \: \: \: \: - - ( {eq}^{n} 1)$

PA = PB (By distance formula)

$\sqrt{ {(x - 3)}^{2} + {(y - 4)}^{2} } = \sqrt{ {(x - 5)}^{2} + {(y + 2)}^{2} } \\ {x}^{2} + 9 - 6x + {y}^{2} + 16 - 8y = {x}^{2} + 25 - 10x + {y}^{2} + 4 + 8y \\ - 6x - 8y = - 10x + 8y + 4 \\ 10x - 6x - 8y - 8y = 4 \\ 4x - 16y = 4 \\ x - 4y = 1 \\ x = 1 + 4y \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - - ( {eq}^{n} \: \: 2)$

Put x = 1 + 4y in eq 1

$3(1 + 4y) + y = 23 \\ 3 + 12y + y = 23 \\ 3 + 13y = 23 \\ 13y = 23 - 3 \\ 13y = 20 \\ y = \frac{20}{13}$

By eq 2

$x = 1 + 4( \frac{20}{13} ) \\ x = 1 + \frac{80}{13} \\ x = \frac{13 + 80}{13} \\ x = \frac{93}{13}$

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