Find the points on X - ax is which is equidistant that from (2,-5) and (-2,9)
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points of X axis = (x1+x2)/2
= [2 +(-2)]/2 = 0/2 = 0
= [2 +(-2)]/2 = 0/2 = 0
Sanjana111111:
(x1 + x2)/2
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It
this is formula and after that i have solved
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HELLO DEAR,
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
=>√[(x2 - x1)² + (y2 - y1)²]
PA = √[(2 - x)² + (-5 - 0)²]
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)
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
=>√[(x2 - x1)² + (y2 - y1)²]
PA = √[(2 - x)² + (-5 - 0)²]
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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