Find the point on x-axis which are equidistant from 7, - 4 and 3, 6
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thus the point is on x axis so y axis point will be 0
let it is A (x,0)
it is equidisant to point (7,-4) and (3,6)
and B ( 7 , -4) and C ( 3,6)
as it is of equal length
AB = AC
√[ ( x - 7)² + ( 0 + 4) ² ] = √ [ ( x -3)² + (0-6) ²]
squaring both sides
( x - 7)² + ( 0 + 4) ² = ( x -3)² + (0-6) ²
x² - 14x + 49 + 16 = x² - 6x + 9 + 36
-14x + 6x = 45 - 16 -49
-8x = -20
x = 20/8
x = 5/2
point is ( 5 /2 , 0)
let it is A (x,0)
it is equidisant to point (7,-4) and (3,6)
and B ( 7 , -4) and C ( 3,6)
as it is of equal length
AB = AC
√[ ( x - 7)² + ( 0 + 4) ² ] = √ [ ( x -3)² + (0-6) ²]
squaring both sides
( x - 7)² + ( 0 + 4) ² = ( x -3)² + (0-6) ²
x² - 14x + 49 + 16 = x² - 6x + 9 + 36
-14x + 6x = 45 - 16 -49
-8x = -20
x = 20/8
x = 5/2
point is ( 5 /2 , 0)
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