Question

find the ratio in which the line segment A(1,-5​

Answer 1

Answer:

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

Given:

Two points:

• A(3,4)

• B(5,-2)

PA = PB

Area of $\triangle$PAB = 10 sq. units

To Find:

The coordinates of P.

Solution:

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

We know that,

Area of a triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

Let the side AB , PB and PA be a , b and c respectively.

Using Distance Formula,

• Distance Formula = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

First, calculating AB.

AB = $\sqrt{(5-3)^2+(-2-4)^2}$

AB = $\sqrt{4+36}$

AB = $\sqrt{40}$

Now, calculating PB.

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

PB = $\sqrt{x^2+25-10x+y^2+4+8y}$

PB = $\sqrt{x^2+y^2-10x+8y+29}$

Now , calculating PA

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

PA = $\sqrt{x^2+9-6x+y^2+16-8y}$

PA = $\sqrt{x^2+y^2-6x-8y+25}$

Since,

PA = PB

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

Squaring both sides, we get:

$x^2+y^2-6x-8y+25-x^2-y^2+10x-8y-29=0$

x-3y = 1

x = 3y+1 ....(1)

We know that,

Semi Perimeter = $\dfrac{a+b+c}{2}$

Here,

a = $\sqrt{40}$

b = c = $\sqrt{x^2+y^2-10x+4y+29}$

Semi Perimeter = $\dfrac{\sqrt{40}+\sqrt{x^2+y^2-10x+4y+29}+\sqrt{x^2+y^2-10x+4y+29}}{2}$

Semi Perimeter = $\cancel{2}\left(\dfrac{\sqrt{10}+\sqrt{x^2+y^2-10x+4y+29}}{\cancel{2}}\right)$

Semi Perimeter = $\sqrt{10}+\sqrt{x^2+y^2-10x+4y+29}$

We know that,

Area of $\triangle$PAB=$\sqrt{\sqrt{10}+\sqrt{x^2+y^2-10x+4y+29}(\sqrt{10}+\sqrt{x^2+y^2-10x+4y+29}-\sqrt{40})(\sqrt{10}+\sqrt{x^2+y^2-10x+4y+29}-\sqrt{x^2+y^2-10x+4y+29})(\sqrt{10}+\sqrt{x^2+y^2-10x+4y+29}-\sqrt{x^2+y^2-10x+4y+29})}$

10 = $\sqrt{\sqrt{10}+\sqrt{x^2+y^2-10x+4y+29}(\sqrt{10}+\sqrt{x^2+y^2-10x+4y+29}-2\sqrt{10})(\sqrt{10})(\sqrt{10})}$

10 = $\sqrt{(\sqrt{x^2+y^2-10x+4y+29}+\sqrt{10})(\sqrt{x^2+y^2-10x+4y+29}-\sqrt{10})(10)}$

We know that,

(a+b)(a-b) = $(a^2-b^2)$

10 = $\sqrt{(x^2+y^2-10x+4y+29+10)(10)}$

Squaring both sides , we get:

100 = ($x^2+y^2-10x+4y+29-10$)10

10 = $x^2+y^2-10x+4y+29-10$

Substituting the value of x from (1),

10 = $(3y+1)^2+y^2-10(3y+1)+4y+19$

$9y^2+y^2+6y+4y-30y-10+1+19-10=0$

$10y^2-20y=0$

10y(y-2)=0

y = 0 or y = 2

Substituting both values of y in (1).

x = 3*0+1 or x = 3*2+1

x = 1 or x = 7.

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