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Find the values of y for which the distance between the point p2,-3 and q 10, y is 10 units

1 Answer

4 votes

Answer:

y = - 9, y = 3

Explanation:

Calculate distance d using the distance formula

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

with (x₁, y₁ ) = (2, - 3) and (x₂, y₂ ) = (10, y)

d =
√((10-2)^2+(y+3)^2)

=
√(8^2+(y+3)^2)

Given distance between points is 10, then


√(64 +(y+3)^2) = 10 ( square both sides )

64 + (y + 3)² = 100 ( subtract 64 from both sides )

(y + 3)² = 36 ( take the square root of both sides )

y + 3 = ±
√(36) = ± 6 ( subtract 3 from both sides )

y = - 3 ± 6 , thus

y = - 3 - 6 = - 9

y = - 3 + 6 = 3

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