Question

Find the value of p, if (0, p) is equidistant from (3, 2) and (-5,-2).​

Answer 1

Answer:

Correct option is

C

5.25

(p−3)

2

+(0−2)

2

=(p−5)

2

+(0−3)

2

or p

2

−6p+9+4=p

2

−10p+25+9

or p=

4

21

=5⋅25

Answer 2

Answer:

-2

Step-by-step explanation:

Given :

(0, p) is equidistant from (3, 2) and (-5,-2).​

To find :

The value of p

Solution :

Let A be point (3, 2) and let B be point (-5,-2)

Let P be the point (0, p)

Now given condition is that, P is equidistant from A and B

So, PA = PB

We know that, distance between two points of coordinates, (x₁, y₁) and (x₂, y₂)

$\boxed{\mathrm{d = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2} }}$

So, squaring on both sides,

=> PA² = PB²

$\implies \mathrm{(0-3)^2 + (p-2)^2 = (0+5)^2+(p+2)^2}$

$\implies \mathrm{9 + p^2 -4p+4 = 25 + p^2+4p+4}$

$\implies \mathrm{9 + \not{p^2} -4p+4 = 25 + \not{p^2}+4p+4}$

$\implies \mathrm{13 -4p = 4p + 29}$

$\implies \mathrm{8p = -16}$

$\implies \mathrm{p = -2}$

$\therefore \underline{\boxed{\mathbf{p = -2}}}$

Thus, the value of p = -2

Hope it helps!