Question

Find the y-coordinates of the points that are 10 units away from the point (-1,6) that have an x-coordinate of 7

Answer 1

Answer:

y = 0 or 12

Step-by-step explanation:

$Let \ (x _ 1 , y _ 1) \ and \ (x _ 2 , y _ 2 ) \ be \ the \ coordinates \ \\\\Distance \ between \ the \ points = \sqrt{(x_2 - x_1)^2 + (y_ 2 -y_1)^2}$

$Given : distance = 10 \ units , (x_1 , y _ 1 ) = ( - 1 , 6 ) \ and \ (x_2 , y _ 2 ) = (7 , y )$

$Substituting \ the \ values \ \\\\10 = \sqrt{(7 -(-1))^2 + (y - 6)^2}\\\\10 = \sqrt{(7 + 1)^2 + ( y - 6 )^2}\\\\10 = \sqrt{8^2 + ( y - 6)^2 }\\\\10 = \sqrt{64 + (y -6)^2 } \\\\Squaring \ both \ sides \\\\10^2 = (\sqrt{64 + (y -6)^2 })^2\\\\100 = 64 + ( y -6)^2\\\\100 - 6 4 = ( y -6)^2 \\\\36 = ( y - 6 )^2\\\\\sqrt{36} = y - 6\\\\\pm 6 = y - 6 \\\\So, \\\\\ y - 6 = 6 => y = 12\\\\\ \ \ \ y - 6 = - 6 => y = 0$

Answer 2

Answer:

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