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Point A (2, p) lies in the first quadrant at a distance of 3 units from point (2, 2). Find the possible value of p.
- a)5
- b)3
- c)2
- d)1
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Point A (2, p) lies in the first quadrant at a distance of 3 units fro...
Applying distance formula,
3 = p - 2
p = 3 + 2 = 5
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Community Answer
Point A (2, p) lies in the first quadrant at a distance of 3 units fro...
To find the possible value of p, we need to use the distance formula between two points. The distance formula is given by:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Given that point A is at a distance of 3 units from point (2, 2), we can substitute the values into the distance formula and solve for p.
Let's consider the coordinates of point A as (2, p) and the coordinates of point (2, 2) as (x1, y1). Plugging these values into the distance formula, we have:
3 = √((2 - 2)^2 + (p - 2)^2)
Simplifying the equation, we get:
3 = √(0^2 + (p - 2)^2)
3 = √(p^2 - 4p + 4)
Squaring both sides of the equation, we have:
9 = p^2 - 4p + 4
Rearranging the equation, we get:
p^2 - 4p + 4 - 9 = 0
p^2 - 4p - 5 = 0
Now, we can solve this quadratic equation for p. Factoring the equation, we have:
(p - 5)(p + 1) = 0
Setting each factor equal to zero, we get two possible values for p:
p - 5 = 0 or p + 1 = 0
p = 5 or p = -1
Since point A lies in the first quadrant, the possible value of p is 5. Therefore, the correct answer is option A.
d = √((x2 - x1)^2 + (y2 - y1)^2)
Given that point A is at a distance of 3 units from point (2, 2), we can substitute the values into the distance formula and solve for p.
Let's consider the coordinates of point A as (2, p) and the coordinates of point (2, 2) as (x1, y1). Plugging these values into the distance formula, we have:
3 = √((2 - 2)^2 + (p - 2)^2)
Simplifying the equation, we get:
3 = √(0^2 + (p - 2)^2)
3 = √(p^2 - 4p + 4)
Squaring both sides of the equation, we have:
9 = p^2 - 4p + 4
Rearranging the equation, we get:
p^2 - 4p + 4 - 9 = 0
p^2 - 4p - 5 = 0
Now, we can solve this quadratic equation for p. Factoring the equation, we have:
(p - 5)(p + 1) = 0
Setting each factor equal to zero, we get two possible values for p:
p - 5 = 0 or p + 1 = 0
p = 5 or p = -1
Since point A lies in the first quadrant, the possible value of p is 5. Therefore, the correct answer is option A.
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Point A (2, p) lies in the first quadrant at a distance of 3 units from point (2, 2). Find the possible value of p.a)5b)3c)2d)1Correct answer is option 'A'. Can you explain this answer?
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