Question

If P = (2, 5), Q = (x, –7) and PQ = 13, what is the value of ‘x’?​

Answer 1

Answer:

7 or -3

Explanation:

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

=169=(x-2)^2+144

=25=(x-2)^2

=x-2=+5 or -5

x=7 or -3

Answer 2

Answer:

Possible value of x are 7 and -3.

Explanation:

Given that, P (2 , 5)  and Q = (x - 7).

Also given that, PQ = 13.

First, we find the distance between P and Q.

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

• Where, the algebraic phrase known as the distance formula, provides the distances between two places in terms of their coordinates.

Step 1:

We have, PQ = 13.

P = (2 , 5) and Q = (x - 7)

Where, $x_2 = x , x_1 = 2 and y_ 2 = -7 , y_1 = 5$

Distance between P and Q = $\sqrt{(x - 2)^2 + (-7 -5)^2}$

⇒ PQ = $\sqrt{x^2 -4x + 4 + (-12)^2}$

⇒ PQ = $\sqrt{x^2 - 4x +4+ 144}$ = $\sqrt{x^2 - 4x + 148}$

Where given that PQ = 13.

Therefore, 13 = $\sqrt{x^2 - 4x + 148}$

Now, squaring both side we get,

⇒ 169 = $x^2 - 4x + 148$

⇒ 169 - 148 = $x^2 - 4x$

⇒ 21 = $x^2 - 4x$

⇒ $x^2 - 4x - 21 = 0$

⇒ $x^2 - 7x + 3x - 21$

⇒ x(x - 7)+ 3(x - 7)

⇒(x - 7)(x + 3)

So, x - 7 = 0 and x + 3 = 0

⇒ x = 7 and -3

Final answer:

Hence, value of x  are  7 and -3.

#SPJ3