Math, asked by eli27, 1 year ago

One day, three friends Akash, Aditya and Manoj......

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Answers

Answered by dhonisuresh0703
13

Answer:

(-1,6)

Step-by-step explanation:

let P be (x,y)

since the distance are equal,

distance AP=distance BP

using distance formula,

root[ (x-1)^2 +(y-2)^2 ]=root [ (x-3)^2 + (y-8)^2 ]

on squaring,

and simplifying we get,

2 x+12 y=70

=>   x+6 y=35    .........1

since area =10 sq.units

area=1/2 [1(8-y)+3(y-2)+x(2-8)] =10

when you simplify the equation,

6 x-2 y=-18          ................2

solving 1 and 2,

we get

x=-1 and y=6

hence the coordinate is (-1,6)

hope it helps.....

Answered by amitnrw
9

The coordinates of P are  ( 5 , 4)   or ( -1 , 6)  and Aakash Distance from Aditya & Manoj  = √20 m

Step-by-step explanation:

Aditya   point A ( 1 , 2)

Manoj   point B  ( 3 , 8)

Aakash Distance from Aditya & Manoj are equal

Let say coordinate of P = (x , y)

=> (x - 1)² + (y -2)² = (x - 3)² + (y - 8)²

=> x² - 2x + 1 + y² -4y + 4 = x² - 6x + 9 + y² - 16y + 64

=> 4x + 12y = 68

=> x + 3y  = 17

Area of Triangle PAB  P(x , y)  A ( 1 , 2) ,  B  ( 3 , 8)

= (1/2)| x(2 - 8)  + 1 (8 - y) + 3(y - 2)|

= (1/2)| -6x + 8 - y + 3y - 6|

= (1/2) | -6x + 2y + 2|

= -3x + y + 1    or    3x - y - 1

-3x + y + 1  = 10

=> -3x + y = 9

       x + 3y  = 17

=> y = 6  , x = - 1

3x - y - 1 =  10

=> 3x- y = 11

     x + 3y  = 17

=>   x = 5  & y = 4

Aakash Distance from Aditya & Manoj  = √ (5 - 1)² + (4-2)² = √(5 - 3)² + (4 - 8)² = √20

or √ (-1 - 1)² + (6-2)² = √(-1 - 3)² + (6 - 8)² = √20

the coordinates of P are  ( 5 , 4)   or ( -1 , 6)

Aakash Distance from Aditya & Manoj  = √20 m

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