Question

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

Answer 1

PQ=10

⇒ (10−2)² +(y+3)²=10

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

Squaring both sides,

y ² +6y−27=0

⇒y²+9y−3y−37=0

⇒y(y+9)−3(y+9)=0

⇒(y+9)(y−3)=0

∴y=−9,3