Multiple choice

Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.

  1. y = 3

  2. y = -9

  3. Both 1 and 2

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

PQ = 10 units $\sqrt{(10-2)^2 + (y+3)^2} = 10 \Rightarrow \sqrt{64+ y^2 + 6y + 9} = 10$.  Squaring both sides: 64 + y2 + 6y + 9 = 100 $\Rightarrow$ y2 + 6y - 27 = 0 y2 + 9y - 3y - 27 = 0     $\Rightarrow$ y(y + 9) - 3(y + 9) = 0 (y + 9) (y - 3) = 0  $\Rightarrow$ y +9 = 0 $\Rightarrow$ y= -9 and y -3 = 0 $\Rightarrow$ y = 3 

AI explanation

Using the distance formula, distance² = (10–2)² + (y+3)² = 64 + (y+3)². Setting this equal to 10² = 100 gives (y+3)² = 36, so y+3 = ±6, meaning y = 3 or y = –9. Both values satisfy the original distance condition, so both are valid solutions.

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