Question

(iv) A(3, 4) and B(5,2)​

Answer 1

Answer:

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proper question

Answer 2

Step-by-step explanation:

let us consider the coordinates of P is (a,b) then,

PA=

(a−3)

2

+(b−4)

2

PB=

(a−5)

2

+(b+2)

2

WKT PA=PB

(a−3)

2

+(b−4)

2

=

(a−5)

2

+(b+2)

2

On squaring both sides we get,

(a−3)

2

+(b−4)

2

=(a−5)

2

+(b+2)

2

By simplifying this we get,

⇒a−3b=1.....(1)

Let area of triangle PAB = 10 sq.units.

⇒

2

1

∣x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)∣=10

⇒∣a(4+2)+3(−2−b)+5(b−4)∣=20

By simplifying this we get,

3a+b=23.....(2)

or

3a+b=3.....(3)

Solution form (1) and (2)

Multiply eqn (2) by 3 we get

9a+3b=69.....(4)

Now add (1) and (4)

10a=70

a=7

substitute a value in (1)

7−3b=1

b=2

Hence (a,b)=(7,2)

Solution form (1) and (3)

Multiply eqn (3) by 3 we get

9a+3b=9.....(5)

Now add (1) and (5)

10a=10

a=1

substitute a value in (1)

1−3b=1

b=0

Hence (a,b)=(1,0)

Finally the coordinates of P are (1,0) or (7,2).