Question

find the value of x for which the distance between the points p(4,-5) and q(12,x) is 10 units​

Answer 1

Answer:

distance between the Points P (4,5) and Q (12,x) is 10 units.

Distance formula:

D=√(x2 − x1)^2 + (y2 − y1)^2

√[(12-4)^2 + (x-5)^2

√(8)^2 + (x-5)^2

10 = √(8)^2 + (x-5)^2

squaring on both sides gives

100 = 64 + (x-5)^2

36 = (x-5)^2

taking square root on both sides gives

x-5 = 6 or -6

so x has two values, one is 11 and the other one is -1

Answer 2

$\sf \pmb{Answer :}$

Answer:

distance between the Points P (4,5) and Q (12,x) is 10 units.

Distance formula:

D=√(x2 − x1)^2 + (y2 − y1)^2

√[(12-4)^2 + (x-5)^2

√(8)^2 + (x-5)^2

10 = √(8)^2 + (x-5)^2

squaring on both sides gives

100 = 64 + (x-5)^2

36 = (x-5)^2

taking square root on both sides gives

x-5 = 6 or -6

so x has two values, one is 11 and the other one is -1