Find the values of y for which the distance between the points P(2.-3) and Q(10,y) is 10 units.
Answers
Answer:it is solved below.
Step-by-step explanation:
Distance between P(2,-3) and Q(10,y) is
PQ=√(10-2)^2+{y-(-3)}^2
=√8^2+(y+3)^2
=√64+(y^2+6y+9)
=√73+y^2+6y
But PQ= 10 Units given
Therefore, 10=√73+y^2+6y
Squaring both sides we get
100=73+y^2+6y
Or,27= y^2+6y
Or,y^2+6y-27=0
Or,y^2+9y-3y-27=0
Or,y(y+9)-3(y+9)=0
Or,(y+9) (y-3)=0
Therefore, y = -9,3.
These two are required values of y.
Step-by-step explanation:
Given : -
- the points p(2,-3) and Q(10,Y) is 10 units
To Find : -
- find the value of Y for which the distance
Solution : -
the formula to calculate the distance between two points (a,b) and (c,d) is given by
dist d = √(a - c)² + (b - d)²
applying it for this ques we have d=10 units
➻ 10 = √(2 - 10)² + ( - 3 - y)²
Squaring both sides :
➻ 100 = (-8)² + (9 + y² + 6y)
➻ 100 = 64 + 9 + y² + 6y
➻ y² + 6y = 100 - 73
➻ y² + 6y - 27 = 0
➻ y² + 9y - 3y - 27 = 0
➻ y (y + 9) - 3 (y + 9) = 0
➻ (y - 3) (y + 9) = 0
➻ y = 3 or y = -9
both values of y are possible unless no other condition is specified in the question.