Question

38] P is a point which is equidistant from A (3, 4) and B (5,-2). Find the co-ordinates of P, if the
area of the angle APB = 10 square units.​

Answer 1

Correct question :-

P is a point which is equidistant from A (3, 4) and B (5,-2). Find the co-ordinates of P, if the area of the ΔAPB is 10 square units.

Solution :-

Given P(x,y) be equidistant from A(3,4) = B(5,-2)

⇒ AP = PB

Using distance formula

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

$\implies \rm \sqrt{x^{2} - 6x + 9 + y^{2} - 8y + 16 } = \sqrt{ {x}^{2} - 10x + 25+ {y}^{2} + 4y + 4 }$

$\implies \rm \sqrt{x^{2} - 6x + y^{2} - 8y + 25 } = \sqrt{ {x}^{2} - 10x + {y}^{2} + 4y + 29 }$

Squaring on both sides

⇒ x² - 6x + y² - 8y + 25 = x² - 10x + y² + 4y + 29

⇒ - 6x - 8y + 25 = - 10x + 4y + 29

⇒ - 6x + 10x - 8y - 4y = 29 - 25

⇒ 4x - 12y = 4

⇒ x - 3y = 1

⇒ x = 1 + 3y -- eq(1)

Area of the Δ APB = 10 sq. units

By using Area of triangle formula

$\implies \rm Area \ of \ triangle = \dfrac{1}{2} \mid x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_3) \mid$

A(3,4) P(x,y) B(5, - 2)

Here,

• x1 = 3
• y1 = 4
• x2 = x
• y2 = y
• x3 = 5
• y3 = - 2

$\implies \rm 10= \dfrac{1}{2} \mid x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \mid$

$\implies \rm 20= 3(y + 2) + x( - 2- 4) + 5(4 - y)$

$\implies \rm 20= 3y + 6 - 6x+ 20-5y$

⇒ 6x + 2y = 6

⇒ 3x + y = 3

⇒ 3(1 + 3y) + y = 3

[ From eq(1) ]

⇒ 3 + 9y + y = 3

⇒ 10y = 0

⇒ y = 0

Substituting y = 0 in eq(1)

⇒ x = 1 + 3y = 1 + 3(0) = 1

Therefore the coordinates of P are (1,0)