The distance between Point P ( 2 , 2 ) and Q ( 5, x ) is 5 cm .
To FinD :-
The value of x .
SolutioN :-
Given that the distance between two points , P(2,2) and Q (5,x) is 5 cm . We need to Find the value of x . We can use here distance formula which can be used ti Find the distance between two points (x1 , y1 ) and (x2 , y2) as ,
Answer:
hope it helps
Step-by-step explanation:
use distance formula
5=√(5-2)²+(x-2)²
5=√(3)²+(x²+4+4x)
5=√9+4+4x+x²
5=√14+4x+x²
squaring on both sides
25= 14 + 4x + x²
x²+4x-11=0
solve the quadratic equation
GiveN :-
To FinD :-
SolutioN :-
Given that the distance between two points , P(2,2) and Q (5,x) is 5 cm . We need to Find the value of x . We can use here distance formula which can be used ti Find the distance between two points (x1 , y1 ) and (x2 , y2) as ,
[tex]\sf:\implies \pink{ Distance =\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}\\\\\sf:\implies 5cm = \sqrt{ (5-2)^2+(x-2)^2}\\\\\sf:\implies (5cm)^2 = (\sqrt{3^2+x^2+4-4x})^2 \\\\\sf:\implies 25cm^2 = 9 + x^2+4-4x \\\\\sf:\implies x^2+-4x +13 -25 = 0 \\\\\sf:\implies x^2-4x -12 =0 \\\\\sf:\implies x^2-6x +2x-12=0\\\\\sf:\implies x(x-6)+2(x-6) = 0 \\\\\sf:\implies(x-6)(x+2) =0\\\\\sf:\implies\underset{\blue{\sf Required \ Answers}}{\underbrace{\boxed{\pink{\frak{ x = 6 , -2}}}}} [/tex]