Question

find the value of y for which distance below the point p (2, -3) and Q (10, y) is 10 ​

Answer 1

Answer:

y=9

Step-by-step explanation:

√(x2-x1)^2 + (y2-y1)^2 =10

√(10-2)^2+{y-(-3)}^2=10

√64+y^2+9-6y=10

√y^2-6y+73=10

squaring both sides,

y^2-6y+73=100

y^2-6y+73-100=0

y^2-6y-27=0

solving by factorisation method,

y^2-(9-3)y-27=0

y^2-9y+3y-27=0

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

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

y= -3 & 9

Answer 2

Answer:

y=-3&9

Step-by-step explanation:

is write answer

hope it helps you