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7.1 P(2,-5), B(2,9)

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7.1 P(2,-5), B(2,9)

Posted by Vaibhav Singh 6 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 5 months ago

Let the point of x-axis be P(x, 0)
Given A(2, -5) and B(-2, 9) are equidistant from P
That is PA = PB
Hence PA² = PB² → (1)
Distance between two points is {tex}\sqrt{[(x_2 - x_1)^2 + (y_2 - y_1)^2]}{/tex}
PA = {tex}\sqrt{[(2 - x)^2 + (-5 - 0)^2]}{/tex}
PA² = 4 - 4x +x² + 25
= x²- 4x + 29
Similarly, PB² = x²+ 4x + 85
Equation (1) becomes
x²- 4x + 29 = x²+ 4x + 85
- 8x = 56
x = -7
Hence the point on x-axis is (-7, 0)

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CBSE > Class 10 > Mathematics