Question

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

Answer 1

Answer:

3 ×1/3 =3×0.333...=0.999...

Let x=0.999...

10x=9.999...

10x-x=9.999...-0.999...

9x=9

x=1

That is so because 1/3 is *not exactly equal to* 0.333333.

0.333333 is an approximation of 1/3 which is correct to 6 decimal places.

Answer 2

Step-by-step explanation:

$x_{1}=2,x_{2}=10,y_{1}=-3,y_{2}=y\\distance=10units\\using\:distance\:formula:\\ =\sqrt{ {(x_{2} - x_{1})}^{2} + {(y_{2} -y_{1} )}^{2} } =10 \\ square\:on\:both\:sides:\\={(x_{2} - x_{1})}^{2} + {(y_{2} -y_{1} )}^{2}={(10)}^{2}\\={(10 - 2)}^{2} + {(y -(-3) )}^{2}=100\\={(8)}^{2} + {(y +3 )}^{2}=100\\=64 + {y }^{2} + 6y + 9=100\\={y }^{2} + 6y + 73 - 100 = 0\\={y }^{2} + 6y -27 = 0\\ = {y}^{2} + 9y - 3y - 27 = 0 \\ = y(y + 9) - 3(y + 9) = 0 \\ = (y - 3)(y + 9) \\ = y =3 , - 9$

HOPE THIS HELPS

MARK ME BRAINLIEST