Question

Q44. If x is a positive integer
such that distance between
the points P (x,2) and Q(3,-6) is
10 units, then x=

Answer 1

Answer:

pq²=(x1-x2)²+(y1-y2)²

(10)²=(x-2)²+(3-(-6))²

100=x²-4x+4+81

100-81-4=x²-4x

15=x²-4x

x²-4x-15=0

d=b²-4ac

=(4)²-4(1)(-15)

=16+60

=76

x=4+2root19/2

x=4-2root19/2

Answer 2

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Answer:

distance between P(x,2) and Q(3,-6) is 10

the formula is,

AB^2 = (x2- x1)^2 - (y1 - y2)^2

here,

AB = 10

x2 = x

x1 = 3

y2 = 2

y1 = -6

so, 10^2 = (x-3)^2 + (2+6)^2

100 = (x-3)^2 + +(8)^2

100 - 64 = (x-3)^2

36 = x^2 - 6x + 9

x^2 - 6x - 27 = 0

x = 9,-3

if we neglect the negative value, then x = 9

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