Question

find the values of y for which the distance between the points p( 2, -3) and q ( 10 ,y ) is 10 units
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Answer 1

Step-by-step explanation:

Distance

pq = 10 units

√(10 - 2)² + (y + 3)² = 10

(8)² + (y + 3)² = (10)²

64 + y² + 6y + 9 = 100

y² + 6y + 73 = 100

y² + 6y - 27 = 0

y² + 9y - 3y - 27 = 0

y(y + 9) - 3(y + 9) = 0

(y - 3) (y + 9) = 0

y = 3 , -9

Answer 2

Answer:

$pq {}^{2} = ( {} 10 - 2) { }^{2} + (y + 3) {}^{2} = 100 \\ 6 4 + y { }^{2} + 9 + 6y = 100 \\ y {}^{2} + 6y - 27 = 0 \\ (y + 9)(y - 3) = 0 \\ y = - 9 \: \: .y = 3$