is there any body to my question????can.....can.....
now complete ........this with explain &take 100 points.....
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Answered by
6
HELLO DEAR,
Let the point of x-axis be P(x, 0)
Given that:-
A(3, 2) and B(-5, -20) are equidistant from P
PA = PB
=> PA² = PB²
Distance between two points is √[(x2 - x1)² + (y2 - y1)²]
=> PA = √[3 - x)² + (2 - 0)²]
=> PA² = 9 + X² + 6X + 4
=> PA² = 13 + X² + 6X
AND,
PB = √[-5 - x)² + (-20 - 0)²]
=> PB² = [25 + X² + 10X + 400]
=> PB² = 425 + X² + 10X
NOW,
PA² = PB²
=> 13 + X² + 6X = 425 + x² + 10x
=> x² - x² = 10x - 6x + 425 - 13
=> 4x = -412
=> x = -412/4
=> x = -103
Hence the point on x-axis is (-103, 0)
I HOPE ITS HELP YOU DEAR,
THANKS
Let the point of x-axis be P(x, 0)
Given that:-
A(3, 2) and B(-5, -20) are equidistant from P
PA = PB
=> PA² = PB²
Distance between two points is √[(x2 - x1)² + (y2 - y1)²]
=> PA = √[3 - x)² + (2 - 0)²]
=> PA² = 9 + X² + 6X + 4
=> PA² = 13 + X² + 6X
AND,
PB = √[-5 - x)² + (-20 - 0)²]
=> PB² = [25 + X² + 10X + 400]
=> PB² = 425 + X² + 10X
NOW,
PA² = PB²
=> 13 + X² + 6X = 425 + x² + 10x
=> x² - x² = 10x - 6x + 425 - 13
=> 4x = -412
=> x = -412/4
=> x = -103
Hence the point on x-axis is (-103, 0)
I HOPE ITS HELP YOU DEAR,
THANKS
great job
mark as brainliest plz
thanks
i will but send Attaches...plz
bro iske liye no attachment
sorry
answer is correct or not?
Rohit sir come on line i will goes to ask one More question i Hope u can only solve it
ya u maybe ....
Answered by
1
Answer:
Let the point of x-axis be P(x, 0)
Given that:-
A(3, 2) and B(-5, -20) are equidistant from P
PA = PB
=> PA² = PB²
Distance between two points is √[(x2 - x1)² + (y2 - y1)²]
=> PA = √[3 - x)² + (2 - 0)²]
=> PA² = 9 + X² + 6X + 4
=> PA² = 13 + X² + 6X
AND,
PB = √[-5 - x)² + (-20 - 0)²]
=> PB² = [25 + X² + 10X + 400]
=> PB² = 425 + X² + 10X
NOW,
PA² = PB²
=> 13 + X² + 6X = 425 + x² + 10x
=> x² - x² = 10x - 6x + 425 - 13
=> 4x = -412
=> x = -412/4
=> x = -103
Hence the point on x-axis is (-103, 0)
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