Question

find the values of y for which the distance between the points ( 2,-3 ) P and Q (10 , y) is 10 units​

Answer 1

Answer:

given above

Step-by-step explanation:

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Answer 2

Answer:

For this, we use distance formula:

D = $\sqrt{(x_{2}-x_{1}) ^{2}+(y_{2}-y_{1}) ^{2} }$

x₁ = 2

x₂ = 10

y₁ = -3

y₂ = y

D = 10

putting the values into the equation:

10 = $\sqrt{(10-2) ^{2}+(y+3) ^{2} }$

10 = $\sqrt{(8) ^{2}+y^2+9+6y} }$

10 = $\sqrt{64+y^2+9+6y} }$

$10^2 = 64+y^2+9+6y$

$y^2+6y=100-64-9$

$y^2+6y=27$

$y^2+6y-27=0$

$y^2+9y-3y-27=0$

$y(y+9)-3(y+9)=0$

$(y-3)(y+9)=0$

y = 3 or,

y = -9

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