Find p such that the distance between the points (2, 3) and (p, 6) is equal to √13 units.
Answers
Step-by-step explanation:
Given :-
The distance between the points (2, 3) and (p, 6) is equal to √13 units.
To find :-
Find the value of p ?
Solution :-
Given points are (2, 3) and (p, 6)
Let (x1, y1)=(2,3)=> x1 = 2 and y1 = 3
Let (x2, y2)=(p,6)=>x2=p and y2 = 6
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
On Substituting these values in the above formula then
=> √[(p-2)²+(6-3)²]
=> √[(p-2)²+3²]
=> √(p²-4p+4+9)
=> √(p²-4p+13) units
According to the given problem
The distance = √13 units
=> √(p²-4p+13) =√13
On squaring both sides then
=> [√(p²-4p+13)]² = [√13]²
=> p²-4p+13 = 13
=> p²-4p = 13-13
=> p²-4p = 0
=> p(p-4) = 0
=> p = 0 or p-4 = 0
=> p = 0 or p = 4
Therefore, p = 0 and 4
Answer :-
The values of p for the given problem are 0 and 4
Used formulae:-
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
Answer:
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