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 triangle APB= 10units​

Answer 1

Solution -

In the above Question, the following information is given -

P is a point which is equidistant from A(3,4) and. B (5,-2) .

The area of the triangle APB= 10units

Now,

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

Coordinates Of Point A = ( 3, 4 )

Coordinates Of Point B = ( 5, -2)

Now,

We have the following conditions -

1. P is a point which is equidistant from A(3,4) and. B (5,-2) .

2. the area of the triangle APB= 10units

From the first condition-

P is a point which is equidistant from A(3,4) and. B (5,-2) .

So, PA = PB

Using the distance formula -

Squaring On Both Sides -

$\sf{ { [PA ] } ^ 2 = { [ PB ] } ^ 2 } \\ \\ \sf{ {(x-3)}^2 + {(y-4)}^2 = {(x-5)}^2 + {(y+2)}^2 } \\ \\ \sf{ Solving , \: the \: {x}^ 2 \: and \: {y}^2 \: terms \: get \: cancelled . } \\ \\ \sf{ We \: thus \: obtain \: the \: following \: Equation \::- } \\ \\ \sf{ x-3y=1}$

2. Area Of ∆ APB -

Points -

Coordinates of the point P = ( x , y )

Coordinates Of Point A = ( 3, 4 )

Coordinates Of Point B = ( 5, -2)

Area -

$| \begin{array}{c c c}</p><p></p><p>x & y \\</p><p></p><p>3& 4 \\</p><p></p><p>5 & -2 \\</p><p>x & y </p><p>\end{array} |$ = 0

$\sf{ => | 4x - 6 + 5y - 3y - 20 + 2x }$ | = 0

$\sf{ => 3x + y = 23 }$

So, we obtain the following equations -

x - 3 y = 1 ......... { 1 }

3x + y = 23 ............ { 2 }

Multiplying { 1 } by 3 -

3x - 9y = 3 .............. { 3 }

3x + y = 23 ............ { 2 }

Subtracting and Solving -

-10 y = -20

=> y = 2

Now, substituting this value of y into any of the equations -

=> x = 7

Coordinates of the point P = ( x , y )

So,

Coordinates Of Point P = ( 7, 2 )

Answer 2

The co-ordinates of point P is (7, 2).

Step-by-step explanation :

Given :

P is a point which is equidistant from A(3,4) and. B (5,-2)

The area of the triangle APB - 10 units​.

To Find :

The co-ordinates of P.

Solution :

Consider the co-ordinates of point P as - (x, y).

So,

⇒ PA = PB

⇒ PA² = PB²

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

⇒ x² + 9 - 6x + y² + 16 - 8y = x² + 25 - 10x  + y²  + 4 + 4y

⇒ - 6x - 8y  + 25 = - 10x + 4y + 29

⇒ 4x - 12y = 4

⇒ x - 3y = 1

_____________[Equation 1 ]

Now, As given : The area of the triangle APB is 10 units​.

So,

$\implies \sf \dfrac{1}{2}\;\bigg[3(-2-y)+5(y-4)+x(4+2)\bigg]=10\\ \\ \\\implies \sf -6-3y+5y-20+4x+2x = 20\\ \\ \\\implies \sf 6x+2y=46\\ \\ \\\implies \sf 3x+y=23$

_____________[Equation 2 ]

Solving Equation 1 & 2 :

So, we get

⇒ x = 7

⇒ y = 2

Hence,

The co-ordinates of point P is (7, 2).