Question

8. 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

Answer:

let the points be

P(2,-3)

Q(10,y)

distance PQ = 10

distance bet 2 points = $\sqrt{(x2-x1)^{2}+( y2-y1)^{2} }$

squaring on both sides

10² = (x2-x1)² + (y2-y1)²

100 = 8² + y² + 3² + 6y

100 - 64-9 = y² + 6y

27 = y² + 6y

y² +6y - 27 = 0

y² + 9y - 3y - 27 = 0

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

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

hence y may be -9 or +3

hope it helps

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