Question

if p=(2,5)q=(x, -7) then find possible values of x pq =13

Answer 1

p = 2,3

q = X,-7

A/q

PQ =13

PQ^2= 13^2

(x - 2)² + (-7- 5)² = 169

⇒ (x - 2)^2 = 169-144

= 25 = 52

or

(x - 2) = ± 5

⇒ x = 7 or -3.

Answer 2

Answer:

The possible values of x are -3 and 7.

Step-by-step explanation:

Given:-

P = $(2, 5)$, Q = $(x, -7)$ and PQ = $13$.

To find:-

The possible values of x.

Step 1 of 1

According to the question,

P = (2, 5)

⇒ $(x_1, y_1)$ = (2, 5)

Q = (x, -7)

⇒ $(x_2, y_2)$ = (x, -7)

Also, PQ = 13

This means the distance between the points P and Q is 13.

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

$(x_2-x_1)^2+(y_2-y_1)^2$ = $PQ^2$

Now,

Substitute all the given values as follows:

$(x-2)^2+(-7-5)^2=13^2$

$(x-2)^2+(-12)^2=169$

Simplify as follows:

$x^2+4-4x+144=169$

$x^2-4x+148-169=0$

$x^2-4x-21=0$

Using the middle-term splitting method, we have

$x^2-7x+3x-21=0$

$x(x-7)+3(x-7)=0$

$(x-7)(x+3)=0$

$x=7$ and $x=-3$

Therefore, the possible values of x are -3 and 7.

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