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

√(10-2)²+(y+3)²=10   (distance formula)

squaring both sides, we get

(10-2)² + (y+3)² = 10²

8² + (y+3)² =10²

64+  y²+6y+9 =100

y²-6y-27=0

y²+3y-9y-27=0

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

(y-9),(y+3)

y=9 or y= -3