Question

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

Answer 1

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