A (3, 4) અને B (5, -2) બે બિંદુઓ આપેલા છે. જો | vedclass

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Question

Given that

$P A=P B$

area $	riangle P A B=10$ sq.unit

Let $p \equiv(x, y)$

$	herefore P A=P B$

$\Rightarrow PA ^2= PB ^2$

$\Righta

Vedclass · Accepted Answer

$(7, 2)$

Vedclass · Answer

Given that

$P A=P B$

area $	riangle P A B=10$ sq.unit

Let $p \equiv(x, y)$

$	herefore P A=P B$

$\Rightarrow PA ^2= PB ^2$

$\Rightarrow( x -3)^2+( y -4)^2=( x -5)^2+( y +2)^2$

$\Rightarrow x ^2-6 x +9+ y ^2-8 y +16= x ^2-10 x +25+ y ^2+4 y +4$

$\Rightarrow 4 x -12 y =4$

$	herefore x-3 y=1 \quad---(1)$

again

$\operatorname{area}(	riangle PAB )=10$

$\Rightarrow \frac{1}{2}\left| x _1\left( y _2- y _3ight)+ x _2\left( y _3- y _1ight)+ x _3\left( y _1- y _2ight)ight|=10$

$\Rightarrow| x (-2-4)+5(4- y )+3( y -(-2))|=20$

$	herefore 6 x-2 y+26=\pm 20$

From the question we get

$\Rightarrow-6 x -2 y +26=20 \quad  \Rightarrow-6 x -2 y +26=-$

$\Rightarrow-6 x-2 y=-6 \quad  \Rightarrow-6 x-2 y=-46$

$\Rightarrow 3 x + y =3--(2)  \Rightarrow 3 x + y =23 \quad--(3)$

From equ" $(1)$ and $(2)$ we have

$x-3 t =1$

$3 x+y=3$

$	herefore x =1$ and $y =0$

from equan $(1)$ and $(3)$, we get

$x-3 y=1$

$3 x+y=23$

$	herefore x =7$ and $	herefore x =2$

$	herefore$ Required 00 - ordinate of point

$P$ are $(7,2)$

$A\ (3, 4)$ અને $B\ (5, -2)$ બે બિંદુઓ આપેલા છે. જો $PA = PB$ અને $\Delta PAB$ નું ક્ષેત્રફળ = $10$ હોય, તો $P$ શોધો.

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