Question

Find the points on the x axis which are at a distance of 2√5 unit from the point (7 -4) solutions

Answer 1

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Answer 2

Step-by-step explanation:

according to the question ,

The distance between these two points is 2√5

also one point lies on X-axis, so the coordinates of the points will be (x, 0)

therefore

using distance formula we have:- ________________________

√(X -- 7) + ( 0 + 4 ) = 2√5. A(x,0) B(7,-4)

x^2 + 49 - 14x + 16 = 20

x^2 - 14x + 65 - 20 = 0

x^2 - 14x + 45 = 0

x^2 - 9x - 5x + 45 =0

x ( x - 9 ) -5 ( x - 9 ) = 0

(x - 5) =0. OR ( x - 9 )= 0

implies that

either x = 5 or 9

therefore coordinates of point become (5,0) or (9,0)