Find the value of y for which the distance between the points p(2 -3) and q(10 y) is 10 units

Question

asked Feb 21, 2021 59.7k views

1 Answer

Artur Eshenbrener · Answer 1 · 2021-02-26T18:13:22+0000

Answer:

Explanation:

Using the distance formula: distance =
$\sqrt{(x_(2) -x_(1))^(2)+(y_(2) -y_(1))^(2) }$

sub it in,

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

10 =
$\sqrt{(8)^(2) +(y+3)^(2)}$

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

10 =
$\sqrt{73+y^(2) +6y}$

square both sides:

100 = 73 +
y^(2) + 6y

y^(2) + 6y = 27

y( y + 6) = 27

y = 27 OR THE OTHER SOLUTION IS y + 6 = 27

y = 21