Question

Find the value of x for which the distance between the points P(4,-5) and Q(12,x) is 10 units.

Answer 1

SOLUTION

Here, we have

PQ= 10

=)(PQ)^2= (10)^2 = 100

$= > (12 - 4) {}^{2} + (x + 5) {}^{2} = 100 \\ = > 8 {}^{2} + (x + 5) {}^{2} = 100 \\ = > {(x + 5)}^{2} = 100 - 64 = 36\\ = > {(x + 5)}^{2} = {6}^{2} \\ = > x + 5 = + - 6 \\ = > x + 5 = 6 \: \: \: \: \: \: \: or \: \: \: \: \: x + 5 = - 6 \\ = > x = 1 \: \: \: \: \: \: \: or \: \: \: \: x = - 11$

Hence, x= -1 or x= -11

hope it helps ✔️

Answer 2

Heya dear!✌☺

Step-by-step explanation:

P(4,-5) and q(12,x)

100=(12-4)²+(x+5)²

100=8²+x²+5²× 2 * x* 5

100=64+x²+25+10x

100 = 89 + x² +10x

100-89=x²+10x

11 = x²+10x

x² +10x -11 = 0

x²+11x-1x-11=0

x(x-11)-1(x+11)

(x-1)(x+11)

x -1 = 0

x = 1

or

x+11 = 0

x = -11

x= 1, -11