Find the values of y for which the distance between the points P(2,-3) and Q(10,y) is 10 units
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hello users.....
we have given that
distance between the points P(2,-3) and Q(10,y) is 10 units
we have to find value of y = ?
solution:-
according to distance formula
PQ = √(10-2)² + {y-(-3)}²
=> 10= √{8² +(y+3)² }
=> 10=√{64 +(y² +9+6y) }
=> 10= √(73 +y² +6y )
=> 100 = 73 +y² +6y
=> y² +6y -27 = 0
now solving this equation
we get ,
=> y² +9y-3y-27= 0
=> y(y+9) -3(y+9) = 0
=> y+9= 0. and y-3= 0
=> y= 3 ,-9 answer
⭐⭐ hope it helps ⭐⭐
we have given that
distance between the points P(2,-3) and Q(10,y) is 10 units
we have to find value of y = ?
solution:-
according to distance formula
PQ = √(10-2)² + {y-(-3)}²
=> 10= √{8² +(y+3)² }
=> 10=√{64 +(y² +9+6y) }
=> 10= √(73 +y² +6y )
=> 100 = 73 +y² +6y
=> y² +6y -27 = 0
now solving this equation
we get ,
=> y² +9y-3y-27= 0
=> y(y+9) -3(y+9) = 0
=> y+9= 0. and y-3= 0
=> y= 3 ,-9 answer
⭐⭐ hope it helps ⭐⭐
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