Question

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Chapter : Coordinate Geometry
Class : 10 (NCRET)
Explain : If the two points (x,7) & (1,15) is 10. Find the value of x.

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Answer 1

hiii
here is your answe
Say (x1,y1) = (x,7)

(x2,y2) = (1,15)

Distance between two points = √[(x2-x1)^2+(y2-y1)^2]

10 = √[(1-x)^2+(15-7)^2

10 = √(1+x^2-2x+64)

10 = √(x^2-2x+65)

Squring to each side.

100 = x^2-2x+65

Bring all terms to one side.

x^2-2x+65-100 = 0

x^2-2x-35 = 0

x^2-7x+5x-35 = 0

x(x-7)+5(x-7) = 0

(x+5)+ (x-7) = 0

x+5 = 0 and x-7 = 0

x = -5 and x = 7

Answer 2

Using the distance formula
10=√(x-1)^2+(15-7)^2
10=√(x-1)^2+64
100=(x-1)^2+64
(x-1)^2=36
→(x-1)=±6

so x=7 or -5