Question

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

Answer 1

Answer:

y = - 9, y = 3

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

   = $\sqrt{8^2+(y+3)^2}$

Given distance between points is 10, then

$\sqrt{64 +(y+3)^2}$ = 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 = ± $\sqrt{36}$ = ± 6 ( subtract 3 from both sides )

y = - 3 ± 6 , thus

y = - 3 - 6 = - 9

y = - 3 + 6 = 3